• The Stone-Čech compactification, which embeds locally compact Hausdorff spaces into compact Hausdorff spaces in a “universal” fashion; and
• The Urysohn metrization theorem, that shows that every second-countable normal Hausdorff space is metrizable.

— 1. The Stone-Čech compactification —

Observe that any dense open subset of a compact Hausdorff space is automatically a locally compact Hausdorff space. We now study the reverse concept:

Definition 1 A compactification of a locally compact Hausdorff space is an embedding (i.e. a homeomorphism between and ) into a compact Hausdorff space such that the image of is an open dense subset of . We will often abuse notation and refer to as the compactification rather than the embedding , when the embedding is obvious from context.

One compactification is finer than another (or is coarser than ) if there exists a continuous map such that ; notice that this map must be surjective and unique, by the open dense nature of . Two compactifications are equivalent if they are both finer than each other.

Example 1 Any compact set can be its own compactification. The real line can be compactified into by using the arctan function as the embedding, or (equivalently) by embedding it into the extended real line . It can also be compactified into the unit circle by using the stereographic projection . Notice that the former embedding is finer than the latter. The plane can similarly be compactified into the unit sphere by the stereographic projection .

Exercise 1 Let be a locally compact Hausdorff space that is not compact. Define the one-point compactification by adjoining one point to , with the topology generated by the open sets of , and the complement (in ) of the compact sets in . Show that (with the obvious embedding map) is a compactification of . Show that the one-point compactification is coarser than any other compactification of .

We now consider the opposite extreme to the one-point compactification:

Definition 2 Let be a locally compact Hausdorff space. A Stone-Čech compactification of is defined as the finest compactification of , i.e. the compactification of which is finer than every other compactification of .

It is clear that the Stone-Čech compactification, if it exists, is unique up to isomorphism, and so one often abuses notation by referring to the Stone-Čech compactification. The existence of the compactification can be established by Zorn’s lemma (see these lecture notes of mine from last year). We shall shortly give several other constructions of the compactification. (All constructions, however, rely at some point on the axiom of choice, or a related axiom.)

The Stone-Čech compactification obeys a useful functorial property:

Exercise 2 Let be locally compact Hausdorff spaces, with Stone-Čech compactifications . Show that every continuous map has a unique continuous extension . (Hint: uniqueness is easy; for existence, look at the closure of the graph in , which compactifies and thus cannot be strictly finer than .) In the converse direction, if is a compactification of such that every continuous map into a compact space can be extended continuously to , show that is the Stone-Čech compactification.

Example 2 From the above exercise, we can define limits for any bounded continuous function on and any . But one for coarser compactifications, one can only take limits for special types of bounded continuous functions; for instance, using the one-point compactification of , need not exist for a bounded continuous function , e.g. or do not exist. The finer the compactification, the more limits can be defined; for instance the two point compactification of allows one to define the limits and for some additional functions (e.g. is well-defined); and the Stone-Čech compactification is the only compactification which allows one to take limits for any bounded continuous function (e.g. is well-defined for all ).

Now we turn to the issue of actually constructing the Stone-Čech compactifications.

Exercise 3 Let be a locally compact Hausdorff space. Let be the space of continuous functions from to the unit interval, let be the space of tuples taking values in the unit interval, with the product topology, and let be the Gelfand transform , and let be the closure of in .

• Show that is a compactification of . ({\emph Hint}: Use Urysohn’s lemma and Tychonoff’s theorem.)
• Show that is the Stone-Čech compactification of . ({\emph Hint}: If is any other compactification of , we can identify as a subset of , and then project to . Meanwhile, we can embed inside by the Gelfand transform.)

Exercise 4 Let be a discrete topological space, let be the Boolean algebra of all subsets of . By Stone’s representation theorem (Theorem 1 from Notes 1), is isomorphic to the clopen algebra of a Stone space .

• Show that is a compactification of .
• Show that is the Stone-Čech compactification of .
• Identify with the space of ultrafilters on . (See this post for further discussion of ultrafilters, and this post for further discussion of the relationship of ultrafilters to the Stone-Čech compactification.)

Exercise 5 Let be a locally compact Hausdorff space, and let be the space of bounded continuous complex-valued functions on .

• Show that is a unital commutative -algebra (see Section 4 of Notes 12).
• By the commutative Gelfand-Naimark theorem (Theorem 14 of Notes 12), is isomorphic as a unital -algebra to for some compact Hausdorff space (which is in fact the spectrum of . Show that is the Stone-Čech compactification of .
• More generally, show that given any other compactification of , that is isomorphic as a unital -algebra to a subalgebra of that contains (the space of continuous functions from to that converge to a limit at ), with as the spectrum of this algebra; thus we have a canonical identification between compactifications and -algebras between and , which correspond to the Stone-Čech compactification and one-point compactification respectively.

Exercise 6 Let be a locally compact Hausdorff space. Show that the dual of is isomorphic as a Banach space to the space of real signed Radon measures on the Stone-Čech compactification , and similarly in the complex case. In particular, conclude that .

Remark 1 The Stone-Čech compactification can be extended from locally compact Hausdorff spaces to the slightly larger class of Tychonoff spaces, which are those Hausdorff spaces with the property that any closed set and point not in can be separated by a continuous function which equals on and zero on . This compactification can be constructed by a modification of the argument used to establish Exercise 3. However, in this case the space is merely dense in its compactification , rather than open and dense.

Remark 2 A cautionary note: in general, the Stone-Čech compactification is almost never sequentially compact. For instance, it is not hard to show that is sequentially closed in . In particular, these compactifications are usually not metrisable.

— 2. Urysohn’s metrisation theorem —

Recall that a topological space is metrisable if there exists a metric on that space which generates the topology. There are various necessary conditions for metrisability. For instance, we have seen that metric spaces must be normal and Hausdorff. In the converse direction, we have

Theorem 3 (Urysohn’s metrisation theorem) Let be a normal Hausdorff space which is second countable. Then is metrisable.

Proof: (Sketch) This will be a variant of the argument in Exercise 3, but with a countable family of continuous functions in place of .

Let be a countable base for . If are in this base with , we can apply Urysohn’s lemma and find a continuous function which equals on and vanishes outside of . Let be the collection of all such functions; this is a countable family. We can then embed in using the Gelfand transform . By modifying the proof of Exercise 3 one can show that this is an embedding. On the other hand, is a countable product of metric spaces and is thus metrisable (e.g. by enumerating as and using the metric ). Since a subspace of a metrisable space is clearly also metrisable, the claim follows.

Recalling that compact metric spaces are second countable (Lemma 4 of Notes 10), thus we have

Corollary 4 A compact Hausdorff space is metrisable if and only if it is second countable.

Of course, non-metrisable compact Hausdorff spaces exist; is a standard example. Uncountable products of non-trivial compact metric spaces, such as , are always non-metrisable. Indeed, we already saw in Notes 10 that is compact but not sequentially compact (and thus not metrisable) when has the cardinality of the continuum; one can use the first uncountable ordinal to achieve a similar result for any uncountable , and then by embedding one can obtain non-metrisability for any uncountable product of non-trivial compact metric spaces, thus complementing the metrisability of countable products of such spaces. Conversely, there also exist metrisable spaces which are not second countable (e.g. uncountable discrete spaces). So Urysohn’s metrisation theorem does not completely classify the metrisable spaces, however it already covers a large number of interesting cases.

The continuation to this course, 245C, will begin on Monday, March 29.  The topics for this course are still somewhat fluid – but I tentatively plan to cover the following topics, roughly in order:

• spaces and interpolation; fractional integration
• The Fourier transform on (a very quick review; this is of course covered more fully in 247A)
• Schwartz functions, and the theory of distributions
• Hausdorff measure
• The spectral theorem (introduction only; the topic is covered in depth in 255A)

I am open to further suggestions for topics that would build upon the 245AB material, which would be of interest to students, and which would not overlap too substantially with other graduate courses offered at UCLA.

In order to approximate rough functions by more continuous ones, one of course needs tools that can generate continuous functions with some specified behaviour. The two basic tools for this are Urysohn’s lemma, which approximates indicator functions by continuous functions, and the Tietze extension theorem, which extends continuous functions on a subdomain to continuous functions on a larger domain. An important consequence of these theorems is the Riesz representation theorem for linear functionals on the space of compactly supported continuous functions, which describes such functionals in terms of Radon measures.

Sometimes, approximation by continuous functions is not enough; one must approximate continuous functions in turn by an even smoother class of functions. A useful tool in this regard is the Stone-Weierstrass theorem, that generalises the classical Weierstrass approximation theorem to more general algebras of functions.

As an application of this theory (and of many of the results accumulated in previous lecture notes), we will present (in an optional section) the commutative Gelfand-Neimark theorem classifying all commutative unital -algebras.

— 1. Urysohn’s lemma —

Let be a topological space. An indicator function in this space will not typically be a continuous function (indeed, if is connected, this only happens when is the empty set or the whole set). Nevertheless, for certain topological spaces, it is possible to approximate an indicator function by a continuous function, as follows.

Lemma 1 (Urysohn’s lemma) Let be a topological space. Then the following are equivalent:

• (i) Every pair of disjoint closed sets in can be separated by disjoint open neighbourhoods , .
• (ii) For every closed set in and every open neighbourhood of , there exists an open set and a closed set such that .
• (iii) For every pair of disjoint closed sets in , there exists a continuous function which equals on and on .
• (iv) For every closed set in and every open neighbourhood of , there exists a continuous function such that for all .

A topological space which obeys any (and hence all) of (i-iv) is known as a normal space; definition (i) is traditionally taken to be the standard definition of normality. We will give some examples of normal spaces shortly.

Proof: The equivalence of (iii) and (iv) is clear, as the complement of a closed set is an open set and vice versa. The equivalence of (i) and (ii) follows similarly.

To deduce (i) from (iii), let be disjoint closed sets, let be as in (iii), and let be the open sets and .

The only remaining task is to deduce (iv) from (ii). Suppose we have a closed set and an open set with . Applying (ii), we can find an open set and a closed set such that

Applying (ii) two more times, we can find more open sets and closed sets such that

Iterating this process, we can construct open sets and closed sets for every dyadic rational in such that for all , and for any .

If we now define , where ranges over dyadic rationals between and , and with the convention that the empty set has sup and inf , one easily verifies that the sets and are open for every real number , and so is continuous as required.

The definition of normality is very similar to the Hausdorff property, which separates pairs of points instead of closed sets. Indeed, if every point in is closed (a property known as the property), then normality clearly implies the Hausdorff property. The converse is not always true, but (as the term suggests) in practice most topological spaces one works with in real analysis are normal. For instance:

Exercise 1 Show that every metric space is normal.

Exercise 2 Let be a Hausdorff space.

• Show that a compact subset of and a point disjoint from that set can always be separated by open neighbourhoods.
• Show that a pair of disjoint compact subsets of can always be separated by open neighbourhoods.
• Show that every compact Hausdorff space is normal.

Exercise 3 Let be the real line with the usual topology , and let be the topology on generated by and the set consisting only of the rationals ; in other words, is the coarsest refinement of the usual topology that makes the set of rationals an open set. Show that is Hausdorff, with every point closed, but is not normal.

The above example was a simple but somewhat artificial example of a non-normal space. One can create more “natural” examples of non-normal Hausdorff spaces (with every point closed), but establishing non-normality becomes more difficult. The following example is due to Stone.

Exercise 4 Let be the space of natural number-valued tuples , endowed with the product topology (i.e. the topology of pointwise convergence).

• Show that is Hausdorff, and every point is closed.
• For , let be the set of all tuples such that for all outside of a countable set, and such that is injective on this finite set (i.e. there do not exist distinct such that ). Show that are disjoint and closed.
• Show that given any open neighbourhood of , there exists disjoint finite subsets of and an injective function such that for any , any such that for all and is identically on , lies in .
• Show that any open neighbourhood of and any open neighbourhood of necessarily intersect, and so is not normal.
• Conclude that with the product topology is not normal.

The property of being normal is a topological one, thus if one topological space is normal, then any other topological space homeomorphic to it is also normal. However, (unlike, say, the Hausdorff property), the property of being normal is not preserved under passage to subspaces:

Exercise 5 Given an example of a subspace of a normal space which is not normal. (Hint: use Exercise 4, possibly after replacing with a homeomorphic equivalent.)

Let be the space of real continuous compactly supported functions on . Urysohn’s lemma generates a large number of useful elements of , in the case when is locally compact Hausdorff:

Exercise 6 Let be a locally compact Hausdorff space, let be a compact set, and let be an open neighbourhood of . Show that there exists such that for all . (Hint: First use the local compactness of to find a neighbourhood of with compact closure; then restrict to this neighbourhood. The closure of is now a compact set; restrict everything to this set, at which point the space becomes normal.)

One consequence of this exercise is that tends to be dense in many other function spaces. We give an important example here:

Definition 2 (Radon measure) Let be a locally compact Hausdorff space that is also -compact, and let be the Borel -algebra. An (unsigned) Radon measure is a unsigned measure with the following properties:

• (Local finiteness) For any compact subset of , is finite.
• (Outer regularity) For any Borel set of , .
• (Inner regularity) For any Borel set of , .

Example 1 Lebesgue measure on is a Radon measure, as is any absolutely continuous unsigned measure , where . More generally, if is Radon and is a finite unsigned measure which is absolutely continuous with respect to , then is Radon. On the other hand, counting measure on is not Radon (it is not locally finite). It is possible to define Radon measures on Hausdorff spaces that are not -compact or locally compact, but the theory is more subtle and will not be considered here. We will study Radon measures more thoroughly in the next section.

Proposition 3 Let be a locally compact Hausdorff space, and let be a Radon measure on . Then for any , is a dense subset in (real-valued) . In other words, every element of can be expressed as a limit (in ) of continuous functions of compact support.

Proof: Since continuous functions of compact support are bounded, and compact sets have finite measure, we see that is a subspace of . We need to show that the closure of this space contains all of .

Let be a Borel set of finite measure. Applying inner and outer regularity, we can find a sequence of compact sets and open sets such that . Applying Exercise 6, we can then find such that . In particular, this implies (by the squeeze theorem) that converges in to (here we use the finiteness of ); thus lies in for any measurable set . By linearity, all simple functions lie in ; taking closures, we see that any function lies in , as desired.

Of course, the real-valued version of the above proposition immediately implies a complex-valued analogue. On the other hand, the claim fails when :

Exercise 7 Let be a locally compact Hausdorff space that is -compact, and let be a Radon measure. Show that the closure of in is , the space of continuous real-valued functions which vanish at infinity (i.e. for every there exists a compact set such that for all ). Thus, in general, is not dense in .

Thus we see that the norm is strong enough to preserve continuity in the limit, whereas the norms are (locally) weaker and permit discontinuous functions to be approximated by continuous ones.

Another important consequence of Urysohn’s lemma is the Tietze extension theorem:

Theorem 4 (Tietze extension theorem) Let be a normal topological space, let be a bounded interval, let be a closed subset of , and let be a continuous function. Then there exists a continuous function which extends , i.e. for all .

Proof: It suffices to find an continuous extension taking values in the real line rather than in , since one can then replace by (note that min and max are continuous operations).

Let be the restriction map . This is clearly a continuous linear map; our task is to show that it is surjective, i.e. to find a solution to the equation for each . We do this by the standard analysis trick of getting an approximate solution to first, and then using iteration to boost the approximate solution to an exact solution.

Let have sup norm , thus takes values in . To solve the problem , we approximate by . By Urysohn’s lemma, we can find a continuous function such that on the closed set and on the closed set . Now, is not quite equal to ; but observe from construction that has sup norm .

Scaling this fact, we conclude that, given any , we can find a decomposition , where and .

Starting with any , we can now iterate this construction to express for all , where and . As is a Banach space, we see that converges absolutely to some limit , and that , as desired.

Remark 1 Observe that Urysohn’s lemma can be viewed the special case of the Tietze extension theorem when is the union of two disjoint closed sets, and is equal to on one of these sets and equal to on the other.

Remark 2 One can extend the Tietze extension theorem to finite-dimensional vector spaces: if is a closed subset of a normal vector space and is bounded and continuous, then one has a bounded continuous extension . Indeed, one simply applies the Tietze extension theorem to each component of separately. However, if the range space is replaced by a space with a non-trivial topology, then there can be topological obstructions to continuous extension. For instance, a map from a two-point set into a topological space is always continuous, but can be extended to a continuous map if and only if and lie in the same path-connected component of . Similarly, if is a map from the unit circle into a topological space , then a continuous extension from to exists if and only if the closed curve is contractible to a point in . These sorts of questions require the machinery of algebraic topology to answer them properly, and are beyond the scope of this course.

There are analogues for the Tietze extension theorem in some other categories of functions. For instance, in the Lipschitz category, we have

Exercise 8 Let be a metric space, let be a subset of , and let be a Lipschitz continuous map with some Lipschitz constant (thus for all ). Show that there exists an extension of which is Lipschitz continuous with the same Lipschitz constant . (Hint: A “greedy” algorithm will work here: pick to be as large as one can get away with (or as small as one can get away with.))

One can also remove the requirement that the function be bounded in the Tietze extension theorem:

Exercise 9 Let be a normal topological space, let be a closed subset of , and let be a continuous map (not necessarily bounded). Then there exists an extension of which is still continuous. (Hint: first “compress” to be bounded by working with, say, (other choices are possible), and apply the usual Tietze extension theorem. There will be some sets in which one cannot invert the compression function, but one can deal with this by a further appeal to Urysohn’s lemma to damp the extension out on such sets.)

There is also a locally compact Hausdorff version of the Tietze extension theorem:

Exercise 10 Let be locally compact Hausdorff, let be compact, and let . Then there exists which extends .

Proposition 3 shows that measurable functions in can be approximated by continuous functions of compact support (cf. Littlewood’s second principle). Another approximation result in a similar spirit is Lusin’s theorem:

Theorem 5 (Lusin’s theorem) Let be a locally compact Hausdorff space that is -compact, and let be a Radon measure. Let be a measurable function supported on a set of finite measure, and let . Then there exists which agrees with outside of a set of measure at most .

Proof: Observe that as is finite everywhere, it is bounded outside of a set of arbitrarily small measure. Thus we may assume without loss of generality that is bounded. Similarly, as is -compact (or by inner regularity), the support of differs from a compact set by a set of arbitrarily small measure; so we may assume that is also supported on a compact set . By Theorem 4, it then suffices to show that is continuous on the complement of an open set of arbitrarily small measure; by outer regularity, we may delete the adjective “open” from the preceding sentence.

As is bounded and compactly supported, lies in for every , and using Proposition 3 and Chebyshev’s inequality, it is not hard to find, for each , a function which differs from by at most outside of a set of measure at most (say). In particular, converges uniformly to outside of a set of measure at most , and is therefore continuous outside this set. The claim follows.

Another very useful application of Urysohn’s lemma is to create partitions of unity.

Lemma 6 (Partitions of unity) Let be a normal topological space, and let be a collection of closed sets that cover . For each , let be an open neighbourhood of , which are finitely overlapping in the sense that each has a neighbourhood that intersects at most finitely many of the . Then there exists a continuous function supported on for each such that for all .

If is locally compact Hausdorff instead of normal, and the are compact, then one can take the to be compactly supported.

Proof: Suppose first that is normal. By Urysohn’s lemma, one can find a continuous function for each which is supported on and equals on the closed set . Observe that the function is well-defined, continuous and bounded below by . The claim then follows by setting .

The final claim follows by using Exercise 6 instead of Urysohn’s lemma.

Exercise 11 Let be a topological space. A function is said to be upper semi-continuous if is open for all real , and lower semi-continuous if is open for all real .

1. Show that an indicator function is upper semi-continuous if and only if is closed, and lower semi-continuous if and only if is open.
2. If is normal, show that a function is upper semi-continuous if and only if for all , and lower semi-continuous if and only if for all , where we write if for all .

— 2. The Riesz representation theorem —

Let be a locally compact Hausdorff space which is also -compact. In Definition 2 we defined the notion of a Radon measure. Such measures are quite common in real analysis. For instance, we have the following result.

Theorem 7 Let be a non-negative finite Borel measure on a compact metric space . Then is a Radon measure.

Proof: As is finite, it is locally finite, so it suffices to show inner and outer regularity. Let be the collection of all Borel subsets of such that

It will then suffice to show that every Borel set lies in (note that as is compact, a subset of is closed if and only if it is compact).

Clearly contains the empty set and the whole set , and is closed under complements. It is also closed under finite unions and intersections. Indeed, given two sets , we can find a sequences , of closed sets and open sets such that and . Since

we have (by monotonicity of ) that

and similarly

and so .

One can also show that is closed under countable disjoint unions and is thus a -algebra. Indeed, given disjoint sets and , pick a closed and open such that ; then

and

for any , and the claim follows from the squeeze test.

To finish the claim it suffices to show that every open set lies in . For this it will suffice to show that is a countable union of closed sets. But as is a compact metric space, it is separable (Lemma 4 from Notes 10), and so has a countable dense subset . One then easily verifies that every point in the open set is contained in a closed ball of rational radius centred at one of the that is in turn contained in ; thus is the countable union of closed sets as desired.

This result can be extended to more general spaces than compact metric spaces, for instance to Polish spaces (provided that the measure remains finite). For instance:

Exercise 12 Let be a locally compact metric space which is -compact, and let be an unsigned Borel measure which is finite on every compact set. Show that is a Radon measure.

When the assumptions of are weakened, then it is possible to find locally finite Borel measures that are not Radon measures, but they are somewhat pathological in nature.

Exercise 13 Let be a locally compact Hausdorff space which is -compact, and let be a Radon measure. Define a set to be a countable union of closed sets, and a set to be a countable intersection of open sets. Show that every Borel set can be expressed as the union of an set and a null set, and as a set with a null subset removed.

If is a Radon measure on , then we can define the integral for every , since assigns every compact set a finite measure. Furthermore, is a linear functional on which is positive in the sense that whenever is non-negative. If we place the uniform norm on , then is continuous if and only if is finite; but we will not use continuity for now, relying instead on positivity.

The fundamentally important Riesz representation theorem for such spaces asserts that this is the only way to generate such linear functionals:

Theorem 8 (Riesz representation theorem for , unsigned version) Let be a locally compact Hausdorff space which is also -compact. Let be a positive linear functional. Then there exists a unique Radon measure on such that .

Remark 3 The -compactness hypothesis can be dropped (after relaxing the inner regularity condition to only apply to open sets, rather than to all sets); but I will restrict attention here to the -compact case (which already covers a large fraction of the applications of this theorem) as the argument simplifies slightly.

Proof: We first prove the uniqueness, which is quite easy due to all the properties that Radon measures enjoy. Suppose we had two Radon measures such that ; in particular, we have

for all . Now let be a compact set, and let be an open neighbourhood of . By Exercise 6, we can find with ; applying this to (1), we conclude that

Taking suprema in and using inner regularity, we conclude that ; exchanging and we conclude that and agree on open sets; by outer regularity we then conclude that and agree on all Borel sets.

Now we prove existence, which is significantly trickier. We will initially make the simplifying assumption that is compact (so in particular ), and remove this assumption at the end of the proof.

Observe that is monotone on , thus whenever .

We would like to define the measure on Borel sets by defining . This does not work directly, because is not continuous. To get around this problem we shall begin by extending the functional to the class of bounded lower semi-continuous non-negative functions. We define for such functions by the formula

(cf. Exercise 11). This definition agrees with the existing definition of in the case when is continuous. Since is finite and is monotone, one sees that is finite (and non-negative) for all . One also easily sees that is monotone on : whenever and , and homogeneous in the sense that for all and . It is also easy to verify the super-additivity property for ; this simply reflects the linearity of on , together with the fact that if and , then .

We now complement the super-additivity property with a countably sub-additive one: if is a sequence, and is such that for all , then .

Pick a small . It will suffice to show that (say) whenever is such that , and denotes a quantity bounded in magnitude by , where is a quantity that is independent of .

Fix . For every , we can find a neighbourhood of such that for all ; we can also find such that . By shrinking if necessary, we see from the lower semicontinuity of the and that we can also ensure that for all and .

By normality, we can find open neighbourhoods of whose closure lies in . The form an open cover of . Since we are assuming to be compact, we can thus find a finite subcover of . Applying Lemma 6, we can thus find a partition of unity , where each is supported on .

Let be such that . Then we can write . If is in this sum, then , and thus (for small enough) , and hence . We can then write

and thus

(here we use the fact that and that the continuous compactly supported function is bounded). Observe that only finitely many summands are non-zero. We conclude that

(here we use that and so is finite). On the other hand, for any and any , the expression

is bounded from above by

since and , this is bounded above in turn by

We conclude that

and the sub-additivity claim follows.

Combining sub-additivity and super-additivity we see that is additive: for .

Now that we are able to integrate lower semi-continuous functions, we can start defining the Radon measure . When is open, we define by

which is well-defined and non-negative since is bounded, non-negative and lower semi-continuous. When is closed we define by complementation:

this is compatible with the definition of on open sets by additivity of , and is also non-negative. The monotonicity of implies monotonicity of : in particular, if a closed set lies in an open set , then .

Given any set , define the outer measure

and the inner measure

thus . We call a set measurable if . By arguing as in the proof of Theorem 7, we see that the class of measurable sets is a Boolean algebra. Next, we claim that every open set is measurable. Indeed, unwrapping all the definitions we see that

Each in this supremum is supported in some closed subset of , and from this one easily verifies that . Similarly, every closed set is measurable. We can now extend to measurable sets by declaring when is measurable; this is compatible with the previous definitions of .

Next, let be a countable sequence of disjoint measurable sets. Then for any , we can find open neighbourhoods of and closed sets in such that and . Using the sub-additivity of on , we have . Similarly, from the additivity of we have . Letting , we conclude that is measurable with . Thus the Boolean algebra of measurable sets is in fact a -algebra, and is a countably additive measure on it. From construction we also see that it is finite, outer regular, and inner regular, and therefore is a Radon measure. The only remaining thing to check is that for all . If is a finite non-negative linear combination of indicator functions of open sets, the claim is clear from the construction of and the additivity of on ; taking uniform limits, we obtain the claim for non-negative continuous functions, and then by linearity we obtain it for all functions.

This concludes the proof in the case when is compact. Now suppose that is -compact. Then we can find a partition of unity into continuous compactly supported functions , with each being contained in the support of finitely many . (Indeed, from -compactness and the locally compact Hausdorff property one can find a nested sequence of compact sets, with each in the interior of , such that . Using Exercise 6, one can find functions that equal on and are supported on ; now take and .) Observe that for all . From the compact case we see that there exists a finite Radon measure such that for all ; setting one can verify (using the monotone convergence theorem) that obeys the required properties.

Remark 4 One can also construct the Radon measure using the Carátheodory extension theorem; this proof of the Riesz representation theorem can be found in many real analysis texts. A third method is to first create the space by taking the completion of with respect to the norm , and then define . It seems to me that all three proofs are about equally lengthy, and ultimately rely on the same ingredients; they all seem to have their strengths and weaknesses, and involve at least one tricky computation somewhere (in the above argument, the most tricky thing is the countable subadditivity of on lower semicontinuous functions). I have yet to find a proof of this theorem which is both clean and conceptual, and would be happy to learn of other proofs of this theorem.

Remark 5 One can use the Riesz representation theorem to provide an alternate construction of Lebesgue measure, say on . Indeed, the Riemann integral already provides a positive linear functional on , which by the Riesz representation theorem must come from a Radon measure, which can be easily verified to assign the value to every interval and thus must agree with Lebesgue measure. The same approach lets one define volume measures on manifolds with a volume form.

Exercise 14 Let be a locally compact Hausdorff space which is -compact, and let be a Radon measure. For any non-negative Borel measurable function , show that

and

Similarly, for any non-negative lower semi-continuous function , show that

Now we consider signed functionals on , which we now turn into a normed vector space using the uniform norm. The key lemma here is the following variant of the Jordan decomposition theorem.

Lemma 9 (Jordan decomposition for functions) Let be a (real) continuous linear functional. Then there exist positive linear functions such that .

Proof: For , we define

Clearly for ; one also easily verifies the homogeneity property and super-additivity property for and . On the other hand, if are such that , then we can decompose for some with and ; for instance we can take and . From this we can complement super-additivity with sub-additivity and conclude that .

Every function in can be expressed as the difference of two functions in . From the additivity and homogeneity of on we may thus extend uniquely to be a linear functional on . Since is bounded on , we see that is also. If we then define , one quickly verifies all the required properties.

Exercise 15 Show that among all possible choices for the functionals appearing in the above lemma, there is a unique choice which is minimal in the sense that for any other functionals obeying the conclusions of the lemma, one has and for all .

Define a signed Radon measure on a -compact, locally compact Hausdorff space to be a signed Borel measure whose positive and negative variations are both Radon. It is easy to see that a signed Radon measure generates a linear functional on as before, and is continuous if is finite. We have a converse:

Exercise 16 (Riesz representation theorem, signed version) Let be a locally compact Hausdorff space which is also -compact, and let be a continuous linear functional. Then there exists a unique signed finite Radon measure such that . (Hint: combine Theorem 8 with Lemma 9.)

The space of signed finite Radon measures on is denoted , or for short.

Exercise 17 Show that the space , with the total variation norm , is a real Banach space, which is isomorphic to the dual of both and its completion , thus

Remark 6 Note that the previous exercise generalises the identifications from previous notes. For compact Hausdorff spaces , we have , and thus . For locally compact Hausdorff spaces that are -compact but not compact, we instead have , where is the Stone-Cech compactification of , which we will discuss in later notes.

Remark 7 One can of course also define complex Radon measures to be those complex finite Borel measures whose real and imaginary parts are signed Radon measures, and define to be the space of all such measures; then one has analogues of the above identifications. We omit the details.

Exercise 18 Let be two locally compact Hausdorff spaces that are also -compact, and let be a continuous map. If is an unsigned finite Radon measure on , show that the pushforward measure on , defined by , is a Radon measure on . What happens for infinite measures? Establish the same fact for signed finite Radon measures.

Let be locally compact Hausdorff and -compact. As is equivalent to the dual of the Banach space , it acquires a weak* topology (see Notes 11), known as the vague topology. A sequence of Radon measures then converges vaguely to a limit if and only if for all .

Exercise 19 Let be Lebesgue measure on the real line (with the usual topology).

• Show that the measures converge vaguely as to the Dirac mass at the origin .
• Show that the measures converge vaguely as to the measure . (Hint: Continuous, compactly supported functions are Riemann integrable.)
• Show that the measures converge vaguely as to the zero measure .

Exercise 20 Let be locally compact Hausdorff and -compact. Show that for every unsigned Radon measure , the map defined by sending to the measure is an isometry, thus can be identified with a subspace of . Show that this subspace is closed in the norm topology, but give an example to show that it need not be closed in the vague topology. Show that , where ranges over all unsigned Radon measures on ; thus one can think of as many ‘s “glued together”.

Exercise 21 Let be a locally compact Hausdorff space which is -compact. Let be a sequence of functions, and let be another function. Show that converges weakly to in if and only if the are uniformly bounded and converge pointwise to .

Exercise 22 Let be a locally compact metric space which is -compact.

• Show that the space of finitely supported measures in is a dense subset of in the vague topology.
• Show that a Radon probability measure in can be expressed as the vague limit of a sequence of discrete (i.e. finitely supported) probability measures.

— 3. The Stone-Weierstrass theorem —

We have already seen how rough functions (e.g. functions) can be approximated by continuous functions. Now we study in turn how continuous functions can be approximated by even more special functions, such as polynomials. The natural topology to work with here is the uniform topology (since uniform limits of continuous functions are continuous).

For non-compact spaces, such as , it is usually not possible to approximate continuous functions uniformly by a smaller class of functions. For instance, the function cannot be approximated uniformly by polynomials on , since is bounded, the only bounded polynomials are the constants, and constants cannot converge to anything other than another constant. On the other hand, on a compact domain such as , one can easily approximate uniformly by polynomials, for instance by using Taylor series. So we will focus instead on compact Hausdorff spaces such as , in which continuous functions are automatically bounded.

The space of (real-valued) polynomials is a subspace of the Banach space . But it is also closed under pointwise multiplication , making an algebra, and not merely a vector space. We can then rephrase the classical Weierstrass approximation theorem as the assertion that is dense in .

One can then ask the more general question of when a sub-algebra of – i.e. a subspace closed under pointwise multiplication – is dense. Not every sub-algebra is dense: the algebra of constants, for instance, will not be dense in when has at least two points. Another example in a similar spirit: given two distinct points in , the space is a sub-algebra of , but it is not dense, because it is already closed, and cannot separate and in the sense that it cannot produce a function that assigns different values to and .

The remarkable Stone-Weierstrass theorem shows that this inability to separate points is the only obstruction to density, at least for algebras with the identity.

Theorem 10 (Stone-Weierstrass theorem, real version) Let be a compact Hausdorff space, and let be a sub-algebra of which contains the constant function and separates points (i.e. for every distinct , there exists at least one in such that . Then is dense in .

Remark 8 Observe that this theorem contains the Weierstrass approximation theorem as a special case, since the algebra of polynomials clearly separates points. Indeed, we will use (a very special case) of the Weierstrass approximation theorem in the proof.

Proof: It suffices to verify the claim for algebras which are closed in the topology, since the claim follows in the general case by replacing with its closure (note that the closure of an algebra is still an algebra).

Observe from the Weierstrass approximation theorem that on any bounded interval , the function can be expressed as the uniform limit of polynomials ; one can even write down explicit formulae for such a , though we will not need such formulae here. Since continuous functions on the compact space are bounded, this implies that for any , the function is the uniform limit of polynomial combinations of . As is an algebra, the lie in ; as is closed; we see that lies in .

Using the identities , , we conclude that is a lattice in the sense that one has whenever .

Now let and . We would like to find such that for all .

Given any two points , we can at least find a function such that and ; this follows since the vector space separates points and also contains the identity function (the case needs to be treated separately). We now use these functions to build the approximant . First, observe from continuity that for every there exists an open neighbourhood of such that for all . By compactness, for any fixed we can cover by a finite number of these . Taking the max of all the associated to this finite subcover, we create another function such that and for all . By continuity, we can find an open neighbourhood of such that for all . Again applying compactness, we can cover by a finite number of the ; taking the min of all the associated to this finite subcover we obtain with for all , and the claim follows.

There is an analogue of the Stone-Weierstrass theorem for algebras that do not contain the identity:

Exercise 23 Let be a compact Hausdorff space, and let be a closed sub-algebra of which separates points but does not contain the identity. Show that there exists a unique such that .

The Stone-Weierstrass theorem is not true as stated in the complex case. For instance, the space of complex-valued functions on the closed unit disk has a closed proper sub-algebra that separates points, namely the algebra of functions in that are holomorphic on the interior of this disk. Indeed, by Cauchy’s theorem and its converse (Morera’s theorem), a function lies in if and only if for every closed contour in , and one easily verifies that this implies that is closed; meanwhile, the holomorphic function separates all points. However, the Stone-Weierstrass theorem can be recovered in the complex case by adding one further axiom, namely that the algebra be closed under conjugation:

Exercise 24 (Stone-Weierstrass theorem, complex version) Let be a compact Hausdorff space, and let be a complex sub-algebra of which contains the constant function , separates points, and is closed under the conjugation operation . Then is dense in .

Exercise 25 Let be the space of trigonometric polynomials on the unit circle , where and the are complex numbers. Show that is dense in (with the uniform topology), and that is dense in (with the topology) for all .

Exercise 26 Let be a locally compact Hausdorff space that is -compact, and let be a sub-algebra of which separates points and contains the identity function. Show that for every function there exists a sequence which converges to uniformly on compact subsets of .

Exercise 27 Let be compact Hausdorff spaces. Show that every function can be expressed as the uniform limit of functions of the form , where and .

Exercise 28 Let be a family of compact Hausdorff spaces, and let be the product space (with the product topology). Let . Show that can be expressed as the uniform limit of continuous functions , each of which only depend on finitely many of the coordinates in , thus there exists a finite subset of and a continuous function such that for all .

One useful application of the Stone-Weierstrass theorem is to demonstrate separability of spaces such as .

Proposition 11 Let be a compact metric space. Then and are separable.

Proof: It suffices to show that is separable. By Lemma 4 of Notes 10, has a countable dense subset . By Urysohn’s lemma, for each we can find a function which equals on and is supported on . The can then easily be verified to separate points, and so by the Stone-Weierstrass theorem, the algebra of polynomial combinations of the in are dense; this implies that the algebra of rational polynomial combinations of the are dense, and the claim follows.

Combining this with the Riesz representation theorem and the sequential Banach-Alaoglu theorem, we obtain

Corollary 12 If is a compact metric space, then the closed unit ball of is sequentially compact in the vague topology.

Combining this with Theorem 7, we conclude a special case of Prokhorov’s theorem:

Corollary 13 (Prokhorov’s theorem, compact case) Let be a compact metric space, and let be a sequence of Borel (hence Radon) probability measures on . Then there exists a subsequence of which converge vaguely to another Borel probability measure .

Exercise 29 (Prokhorov’s theorem, non-compact case) Let be a locally compact metric space which is -compact, and let be a sequence of Borel probability measures. We assume that the sequence is tight, which means that for every there exists a compact set such that for all . Show that there is a subsequence of which converges vaguely to another Borel probability measure . If tightness is not assumed, show that there is a subsequence which converges vaguely to a non-negative Borel measure , but give an example to show that this measure need not be a probability measure.

This theorem can be used to establish Helly’s selection theorem:

Exercise 30 (Helly’s selection theorem) Let be a sequence of functions whose total variation is uniformly bounded in , and which is bounded at one point (i.e. is bounded). Show that there exists a subsequence of which converges pointwise almost everywhere on compact subsets of . (Hint: one can deduce this from Prokhorov’s theorem using the fundamental theorem of calculus for functions of bounded variation.)

— 4. The commutative Gelfand-Naimark theorem (optional) —

One particularly beautiful application of the machinery developed in the last few notes is the commutative Gelfand-Naimark theorem, that classifies commutative -algebras, and is of importance in spectral theory, operator algebras, and quantum mechanics.

Definition 14 A complex Banach algebra is a complex Banach space which is also a complex algebra, such that for all . An algebra is unital if it contains a multiplicative identity , and commutative if for all . A -algebra is a complex Banach algebra with an anti-homomorphism map from to (thus , , and for and ) which is an isometry (thus for all ), an involution (thus for all ), and obeys the identity for all .

A homomorphism between two -algebras is a continuous algebra homomorphism such that for all . An isomorphism is an homomorphism whose inverse exists and is also a homomorphism; two -algebras are isomorphic if there exists an isomorphism between them.

Exercise 31 If is a Hilbert space, and is the algebra of bounded linear operators on this space, with the adjoint map and the operator norm, show that is a unital -algebra (not necessarily commutative). Indeed, one can think of -algebras as an abstraction of a space of bounded linear operators on a Hilbert space (this is basically the content of the non-commutative Gelfand-Naimark theorem, which we will not discuss here).

Exercise 32 If is a compact Hausdorff space, show that is a unital commutative -algebra, with involution .

The remarkable (unital commutative) Gelfand-Naimark theorem asserts the converse statement to Exercise 32:

Theorem 15 (Unital commutative Gelfand-Naimark theorem) Every unital commutative -algebra is isomorphic to for some compact Hausdorff space .

There are analogues of this theorem for non-unital or non-commutative -algebras, but for simplicity we shall restrict attention to the unital commutative case. We first need some spectral theory.

Exercise 33 Let be a unital Banach algebra. Show that if is such that , then is invertible. (Hint: use Neumann series.) Conclude that the space of invertible elements of is open.

Define the spectrum of an element to be the set of all such that is not invertible.

Exercise 34 If is a unital Banach algebra and , show that is a compact subset of that is contained inside the disk .

Exercise 35 (Beurling-Gelfand spectral radius formula) If is a unital Banach algebra and , show that is non-empty with . (Hint: To get the upper bound, observe that if is invertible for some , then so is , then use Exercise 34. To get the lower bound, first observe that for any , the function is holomorphic on the complement of , which is already enough (with Liouville’s theorem) to show that is non-empty. Let be arbitrary, then use Laurent series to show that for all and some independent of . Then divide by and use the uniform boundedness principle to conclude.)

Exercise 36 (-algebra spectral radius formula) Let be a unital -algebra. Show that

for all and . Conclude that any homomorphism between -algebras has operator norm at most . Also conclude that

when is self-adjoint.

The next important concept is that of a character.

Definition 16 Let be a unital commutative -algebra. A character of is be an element in the dual Banach space such that , , and for all ; equivalently, a character is a homomorphism from to (viewed as a (unital) algebra). We let be the space of all characters; this space is known as the spectrum of .

Exercise 37 If is a unital commutative -algebra, show that is a compact Hausdorff subset of in the weak-* topology. (Hint: first use the spectral radius formula to show that all characters have operator norm , then use the Banach-Alaoglu theorem.)

Exercise 38 Define an ideal of a unital commutative -algebra to be a proper subspace of such that for all and . Show that if , then the kernel is a maximal ideal in ; conversely, if is a maximal ideal in , show that is closed, and there is exactly one such that . Thus the spectrum of can be canonically identified with the space of maximal ideals in .

Exercise 39 Let be a compact Hausdorff space, and let be the -algebra . Show that for each , the operation is a character of . Show that the map is a homeomorphism from to ; thus the spectrum of can be canonically identified with . (Hint: use Exercise 23 to show the surjectivity of , Urysohn’s lemma to show injectivity, and Corollary 2 of Notes 10 to show the homeomorphism property.)

Inspired by the above exercise, we define the Gelfand representation , by the formula .

Exercise 40 Show that if is a unital commutative -algebra, then the Gelfand representation is a homomorphism of -algebras.

Exercise 41 Let be a non-invertible element of a unital commutative -algebra . Show that vanishes at some . (Hint: the set is a proper ideal of , and thus by Zorn’s lemma is contained in a maximal ideal.)

Exercise 42 Show that if is a unital commutative -algebra, then the Gelfand representation is an isometry. (Hint: use Exercise 36 and Exercise 41.)

Exercise 43 Use the complex Stone-Weierstrass theorem and Exercises 40, 42 to conclude the proof of Theorem 15.

Title: Give yourself an epsilon of room.

Quick description: You want to prove some statement about some object (which could be a number, a point, a function, a set, etc.).  To do so, pick a small , and first prove a weaker statement (which allows for “losses” which go to zero as ) about some perturbed object .  Then, take limits .  Provided that the dependency and continuity of the weaker conclusion on are sufficiently controlled, and is converging to in an appropriately strong sense, you will recover the original statement.

One can of course play a similar game when proving a statement about some object , by first proving a weaker statement on some approximation to for some large parameter N, and then send at the end.

General discussion: Here are some typical examples of a target statement , and the approximating statements that would converge to :

	for some independent of
is finite	is bounded uniformly in
for all (i.e. maximises f)	for all (i.e. nearly maximises f)
converges as	fluctuates by at most o(1) for sufficiently large n
is a measurable function	is a measurable function converging pointwise to
is a continuous function	is an equicontinuous family of functions converging pointwise to OR is continuous and converges (locally) uniformly to
The event holds almost surely	The event holds with probability 1-o(1)
The statement holds for almost every x	The statement holds for x outside of a set of measure o(1)

Of course, to justify the convergence of to , it is necessary that converge to (or converge to , etc.) in a suitably strong sense. (But for the purposes of proving just upper bounds, such as , one can often get by with quite weak forms of convergence, thanks to tools such as Fatou’s lemma or the weak closure of the unit ball.)  Similarly, we need some continuity (or at least semi-continuity) hypotheses on the functions f, g appearing above.

It is also necessary in many cases that the control on the approximating object is somehow “uniform in “, although for “-closed” conclusions, such as measurability, this is not required. [It is important to note that it is only the final conclusion on that needs to have this uniformity in ; one is permitted to have some intermediate stages in the derivation of that depend on in a non-uniform manner, so long as these non-uniformities cancel out or otherwise disappear at the end of the argument.]

By giving oneself an epsilon of room, one can evade a lot of familiar issues in soft analysis.  For instance, by replacing “rough”, “infinite-complexity”, “continuous”,  “global”, or otherwise “infinitary” objects with “smooth”, “finite-complexity”, “discrete”, “local”, or otherwise “finitary” approximants , one can finesse most issues regarding the justification of various formal operations (e.g. exchanging limits, sums, derivatives, and integrals).  [It is important to be aware, though, that any quantitative measure on how smooth, discrete, finite, etc. should be expected to degrade in the limit , and so one should take extreme caution in using such quantitative measures to derive estimates that are uniform in .]  Similarly, issues such as whether the supremum of a function on a set is actually attained by some maximiser become moot if one is willing to settle instead for an almost-maximiser , e.g. one which comes within an epsilon of that supremum M (or which is larger than , if M turns out to be infinite).  Last, but not least, one can use the epsilon room to avoid degenerate solutions, for instance by perturbing a non-negative function to be strictly positive, perturbing a non-strictly monotone function to be strictly monotone, and so forth.

To summarise: one can view the epsilon regularisation argument as a “loan” in which one borrows an epsilon here and there in order to be able to ignore soft analysis difficulties, and can temporarily be able to utilise estimates which are non-uniform in epsilon, but at the end of the day one needs to “pay back” the loan by establishing a final “hard analysis” estimate which is uniform in epsilon (or whose error terms decay to zero as epsilon goes to zero).

A variant: It may seem that the epsilon regularisation trick is useless if one is already in “hard analysis” situations when all objects are already “finitary”, and all formal computations easily justified.  However, there is an important variant of this trick which applies in this case: namely, instead of sending the epsilon parameter to zero, choose epsilon to be a sufficiently small (but not infinitesimally small) quantity, depending on other parameters in the problem, so that one can eventually neglect various error terms and to obtain a useful bound at the end of the day.  (For instance, any result proven using the Szemerédi regularity lemma is likely to be of this type.)  Since one is not sending epsilon to zero, not every term in the final bound needs to be uniform in epsilon, though for quantitative applications one still would like the dependencies on such parameters to be as favourable as possible.

Prerequisites: Graduate real analysis.  (Actually, this isn’t so much a prerequisite as it is a corequisite: the limiting argument plays a central role in many fundamental results in real analysis.)  Some examples also require some exposure to PDE.

Example 0. The “soft analysis” components of any real analysis textbook will contain a large number of examples of this trick in action.  In particular, any argument which exploits Littlewood’s three principles of real analysis is likely to utilise this trick.  Of course, this trick will also occur repeatedly in my 245B lecture notes. 

Example 1. (Riemann-Lebesgue lemma)  Given any absolutely integrable function , the Fourier transform is defined by the formula

.

The Riemann-Lebesgue lemma asserts that as .  It is difficult to prove this estimate for f directly, because this function is too “rough”: it is absolutely integrable (which is enough to ensure that exists and is bounded), but need not be continuous, differentiable, compactly supported, bounded, or otherwise “nice”.  But suppose we give ourselves an epsilon of room.  Then, as the space of test functions is dense in (a fact I will prove later in this course), we can approximate to any desired accuracy in the norm by a smooth, compactly supported function , thus

. (1)

The point is that is much better behaved than f, and it is not difficult to show the analogue of the Riemann-Lebesgue lemma for .  Indeed, being smooth and compactly supported, we can now justifiably integrate by parts to obtain

for any non-zero , and it is now clear (since is bounded and compactly supported) that as .

Now we need to take limits as .  It will be enough to have converge uniformly to .  But from (1) and the basic estimate

(2)

(which is the single “hard analysis” ingredient in the proof of the lemma) applied to , we see (by the linearity of the Fourier transform) that

and we obtain the desired uniform convergence. 

Remark 1. The same argument also shows that is continuous; we leave this as an exercise to the reader. 

Remark 2. Example 1 is a model case of a much more general instance of the limiting argument: in order to prove a convergence or continuity theorem for all “rough” functions in a function space, it suffices to first prove convergence or continuity for a dense subclass of “smooth” functions, and combine that with some quantitative estimate in the function space (in this case, (2)) in order to justify the limiting argument.

Example 2. The limiting argument in Example 1 relied on the linearity of the Fourier transform .  But, with more effort, it is also possible to extend this type of argument to nonlinear settings.  We will sketch (omitting several technical details, which can be found for instance in my PDE book) a very typical instance.  Consider a nonlinear PDE, e.g. the nonlinear wave equation

(3)

where is some scalar field, and the t and x subscripts denote differentiation of the field .  Formally – if u is sufficiently smooth, and sufficiently decaying at spatial infinity, one can show that the energy

(4)

is conserved, thus for all t.  This can be formally justified by computing the derivative by differentiating under the integral sign, integrating by parts, and then applying the PDE (3); we leave this as an exercise for the reader.  (There are also more fancy ways to see why the energy is conserved, using Hamiltonian or Lagrangian mechanics or by the more general theory of stress-energy tensors, but we will not discuss these here.)  However, these justifications do require a fair amount of regularity on the solution u; for instance, requiring u to be three-times continuously differentiable in space and time, and compactly supported in space on each bounded time interval, would be sufficient to make the computations rigorous by applying “off the shelf” theorems about differentiation under the integration sign, etc.

But suppose one only has a much rougher solution, for instance an energy class solution which has finite energy (4), but for which higher derivatives of u need not exist in the classical sense. (There is a non-trivial issue regarding how to make sense of the PDE (3) when u is only in the energy class, since the terms and do not then make sense classically, but there are standard ways to deal with this, e.g. using weak derivatives, which we will not discuss further here.)  Then it is difficult to justify the energy conservation law directly.  However, it is still possible to obtain energy conservation by the limiting argument.  Namely, one takes the energy class solution u at some initial time (e.g. t=0) and approximates that initial data (the initial position and initial data ) by a much smoother (and compactly supported) choice of initial data, which converges back to in a suitable “energy topology” related to (4) which we will not define here.  It then turns out (from the existence theory of the PDE (3)) that one can extend the smooth initial data to other times t, providing a smooth solution to that data.  For this solution, the energy conservation law can be justified.

Now we take limits as (keeping t fixed).  Since converges in the energy topology to , and the energy functional E is continuous in this topology, converges to .  To conclude the argument, we will also need to converge to , which will be possible if converges in the energy topology to .  Thus in turn follows from a fundamental fact (which requires a certain amount of effort to prove) about the PDE to (4), namely that it is well-posed in the energy class.  This means that not only do solutions exist and are unique for initial data in the energy class, but they depend continuously on the initial data in the energy topology; small perturbations in the data lead to small perturbations in the solution, or more formally that the map from data to solution (say, at some fixed time t) is continuous in the energy topology.  This final fact concludes the limiting argument and gives us the desired conservation law .

Remark 3. It is important that one have a suitable well-posedness theory in order to make the limiting argument work for rough solutions to a PDE; without such a well-posedness theory, it is possible for quantities which are formally conserved to cease being conserved when the solutions become too rough or otherwise “weak”; energy, for instance, could disappear into a singularity and not come back.  

Example 3. (Maximum principle)  The maximum principle is a fundamental tool in elliptic and parabolic PDE (for example, it is used heavily in the proof of the Poincaré conjecture, see e.g. my lecture notes on this topic).  Here is a model example of this principle:

Proposition 1. Let be a smooth harmonic function on the closed unit disk .  If M is a bound such that on the boundary .  Then on the interior as well.

A naive attempt to prove Proposition 1 comes very close to working, and goes like this:  suppose for contradiction that the proposition failed, thus u exceeds M somewhere in the interior of the disk.  Since u is continuous, and the disk is compact, there must then be a point in the interior of the disk where the maximum is attained.  Undergraduate calculus then tells us that and are non-positive, which almost contradicts the harmonicity hypothesis .  However, it is still possible that and both vanish at , so we don’t yet get a contradiction.

But we can finish the proof by giving ourselves an epsilon of room.  The trick is to work not with the function u directly, but with the modified function , to boost the harmonicity into subharmonicity.  Indeed, we have .  The preceding argument now shows that cannot attain its maximum in the interior of the disk; since it is bounded by on the boundary of the disk, we we conclude that is bounded by on the interior of the disk as well.  Sending we obtain the claim. 

Remark 4. Of course, Proposition 1 can also be proven by much more direct means, for instance via the Green’s function for the disk.  However, the argument given is extremely robust and applies to a large class of both linear and nonlinear elliptic and parabolic equations, including those with rough variable coefficients.

Exercise 1.  Use the maximum modulus principle to prove the Phragmén-Lindelöf principle: if f is complex analytic on the strip , is bounded in magnitude by 1 on the boundary of this strip, and obeys a growth condition on the interior of the strip, then show that f is bounded in magnitude by 1 throughout the strip.   (Hint: multiply f by for some even integer m.)

Example 4. (Manipulating generalised functions) In PDE one is primarily interested in smooth (classical) solutions; but for a variety of reasons it is useful to also consider rougher solutions.  Sometimes, these solutions are so rough that they are no longer functions, but are measures, distributions, or some other concept of “generalised function” or “generalised solution“.  For instance, the fundamental solution to a PDE is typically just a distribution or measure, rather than a classical function.  A typical example: a (sufficiently smooth) solution to the three-dimensional wave equation with initial position and initial velocity is given by the classical formula

where is the unique rotation-invariant probability measure on the sphere of radius t, or equivalently, the area element dS on that sphere divided by the surface area of that sphere.  (The convolution of a smooth function and a (compactly supported) finite measure is defined by .)

For this and many other reasons, it is important to manipulate measures and distributions in various ways.  For instance, in addition to convolving functions with measures, it is also useful to convolve measures with measures; the convolution of two finite measures on is defined as the measure which assigns to each measurable set E in , the measure

. (5)

For sake of concreteness, let’s focus on a specific question, namely to compute (or at least estimate) the measure , where is the normalised rotation-invariant measure on the unit circle .  It turns out that while is not absolutely continuous with respect to Lebesgue measure , the convolution is: for some absolutely integrable function f on .  But what is this function f?  It certainly is possible to compute it from the definition (5), or by other methods (e.g. the Fourier transform), but I would like to give one approach to computing these sorts of expressions involving measures (or other generalised functions) based on epsilon regularisation, which requires a certain amount of geometric computation but which I find to be rather visual and conceptual, compared to more algebraic approaches (e.g. based on Fourier transforms).  The idea is to approximate a singular object, such as the singular measure , by a smoother object , such as an absolutely continuous measure.  For instance, one can approximate by

where .  It is clear that converges to in the vague topology, which implies that converges to in the vague topology also.  Since

,

we will be able to understand the limit f by first considering the function

and then taking (weak) limits as to recover f.

Up to constants, one can compute from elementary geometry that is comparable to , and vanishes for , and is comparable to for (and of size in the transition region ) and is comparable to for (and of size about when .  (This is a good exercise for anyone who wants practice in quickly computing the orders of magnitude of geometric quantities such as areas; for such order of magnitude calculations, quick and dirty geometric methods tend to work better here than the more algebraic calculus methods you would have learned as an undergraduate.) The bounds here are strong enough to allow one to take limits and conclude what f looks like: it is comparable to .  And by being more careful with the computations of area, one can compute the exact formula for f(x), though I will not do so here.

Remark 5. Epsilon regularisation also sheds light on why certain operations on measures or distributions are not permissible.  For instance, squaring the Dirac delta function will not give a measure or distribution, because if one looks at the squares of some smoothed out approximations to the Dirac function (i.e. approximations to the identity), one sees that their masses go to infinity in the limit , and so cannot be integrated against test functions uniformly in .  On the other hand, derivatives of the delta function, while no longer measures (the total variation of derivatives of become unbounded), are at least still distributions (the integrals of derivatives of against test functions remain convergent). 

However, in some cases it is useful to work in topologies on vector spaces that are weaker than a norm topology. One reason for this is that many important modes of convergence, such as pointwise convergence, convergence in measure, smooth convergence, or convergence on compact subsets, are not captured by a norm topology, and so it is useful to have a more general theory of topological vector spaces that contains these modes. Another reason (of particular importance in PDE) is that the norm topology on infinite-dimensional spaces is so strong that very few sets are compact or pre-compact in these topologies, making it difficult to apply compactness methods in these topologies. Instead, one often first works in a weaker topology, in which compactness is easier to establish, and then somehow upgrades any weakly convergent sequences obtained via compactness to stronger modes of convergence (or alternatively, one abandons strong convergence and exploits the weak convergence directly). Two basic weak topologies for this purpose are the weak topology on a normed vector space , and the weak* topology on a dual vector space . Compactness in the latter topology is usually obtained from the Banach-Alaoglu theorem (and its sequential counterpart), which will be a quick consequence of the Tychonoff theorem (and its sequential counterpart) from the previous lecture.

The strong and weak topologies on normed vector spaces also have analogues for the space of bounded linear operators from to , thus supplementing the operator norm topology on that space with two weaker topologies, which (somewhat confusingly) are named the strong operator topology and the weak operator topology.

— 1. Topological vector spaces —

We begin with the definition of a topological vector space, which is a space with suitably compatible topological and vector space structures on it.

Definition 1 A topological vector space is a real or complex vector space , together with a topology such that the addition operation and the scalar multiplication operation or is jointly continuous in both variables (thus, for instance, is continuous from with the product topology to ).

It is an easy consequence of the definitions that the translation maps for and the dilation maps for non-zero scalars are homeomorphisms on ; thus for instance the translation or dilation of an open set (or a closed set, a compact set, etc.) is open (resp. closed, compact, etc.). We also have the usual limit laws: if and in a topological vector space, then , and if in the field of scalars, then . (Note how we need joint continuity here; if we only had continuity in the individual variables, we could only conclude that (for instance) if one of or was constant.)

We now give some basic examples of topological vector spaces.

Exercise 1 Show that every normed vector space is a topological vector space, using the balls as the base for the topology. Show that the same statement holds if the vector space is quasi-normed rather than normed.

Exercise 2 Every semi-normed vector space is a topological vector space, again using the balls as a base for the topology. This topology is Hausdorff if and only if the semi-norm is a norm.

Example 1 Any linear subspace of a topological vector space is again a topological vector space (with the induced topology).

Exercise 3 Let be a vector space, and let be a (possibly infinite) family of topologies on , each of which turning into a topological vector space. Let be the topology generated by (i.e. it is the weakest topology that contains all of the . Show that is also a topological vector space. Also show that a sequence converges to a limit in if and only if in for all . (The same statement also holds if sequences are replaced by nets.) In particular, by Exercise 2, we can talk about the topological vector space generated by a family of semi-norms on .

Exercise 4 Let be a linear map between vector spaces. Suppose that we give the topology induced by a family of semi-norms , and the topology induced by a family of semi-norms . Show that is continuous if and only if, for each , there exists a finite subset of and a constant such that for all .

Example 2 (Pointwise convergence) Let be a set, and let be the space of complex-valued functions ; this is a complex vector space. Each point gives rise to a seminorm . The topology generated by all of these seminorms is the topology of pointwise convergence on (and is also the product topology on this space); a sequence converges to in this topology if and only if it converges pointwise. Note that if has more than one point, then none of the semi-norms individually generate a Hausdorff topology, but when combined together, they do.

Example 3 (Uniform convergence) Let be a topological space, and let be the space of complex-valued continuous functions . If is not compact, then one does not expect functions in to be bounded in general, and so the sup norm does not necessarily make into a normed vector space. Nevertheless, one can still define “balls” in by

and verify that these form a base for a topological structure on the vector space, although it is not quite a topological vector space structure because multiplication is no longer continuous. A sequence converges in this topology to a limit if and only if converges uniformly to , thus is finite for sufficiently large and converges to zero as .

Example 4 (Uniform convergence on compact sets) Let and be as in the previous example. For every compact subset of , we can define a seminorm on by . The topology generated by all of these seminorms (as ranges over all compact subsets of ) is called the topology of uniform convergence on compact sets; it is stronger than the topology of poitnwise convergence but weaker than the topology of uniform convergence. Indeed, a sequence converges to in this topology if and only if converges uniformly to on each compact set.

Exercise 5 Show that an arbitrary product of topological vector spaces (endowed with the product topology) is again a topological vector space. [I am not sure if the same statement is true for the box topology; I believe it is false.]

Exercise 6 Show that a topological vector space is Hausdorff if and only if the origin is closed. (Hint: first use the continuity of addition to prove the lemma that if is an open neighbourhood of , then there exists another open neighbourhood of such that , i.e. for all .)

Example 5 (Smooth convergence) Let be the space of smooth functions . One can define the norm on this space for any non-negative integer by the formula

where is the derivative of . The topology generated by all the norms for is the smooth topology: a sequence converges in this topology to a limit if converges uniformly to for each .

Exercise 7 (Convergence in measure) Let be a measure space, and let be the space of measurable functions . Show that the sets

for , , form the base for a topology that turns into a topological vector space, and that a sequence converges to a limit in this topology if and only if it converges in measure.

Exercise 8 Let be given the usual Lebesgue measure. Show that the vector space cannot be given a topological vector space structure in which a sequence converges to in this topology if and only if it converges almost everywhere. (Hint: construct a sequence in which does not converge pointwise a.e. to zero, but such that every subsequence has a further subsequence that converges a.e. to zero, and use Exercise 7′ from Notes 8.) Thus almost everywhere convergence is not “topologisable” in general.

Exercise 9 (Algebraic topology) Recall that a subset of a real vector space is algebraically open if the sets are open for all .

• (i) Show that any set which is open in a topological vector space, is also algebraically open.
• (ii) Give an example of a set in which is algebraically open, but not open in the usual topology. (Hint: a line intersects the unit circle in at most two points.)
• (iii) Show that the collection of algebraically open sets in is a topology.
• (iv) Show that the collection of algebraically open sets in does not give the structure of a topological vector space.

Exercise 10 (Quotient topology) Let be a topological vector space, and let be a subspace of . Let be the space of cosets of ; this is a vector space. Let be the coset map . Show that the collection of sets such that is open gives the structure of a topological vector space. If is Hausdorff, show that is Hausdorff if and only if is closed in .

Some (but not all) of the concepts that are definable for normed vector spaces, are also definable for the more general category of topological vector spaces. For instance, even though there is no metric structure, one can still define the notion of a Cauchy sequence in a topological vector space: this is a sequence such that as (or more precisely, for any open neighbourhood of , there exists such that for all ). It is then possible to talk about a topological vector space being complete (i.e. every Cauchy sequence converges). (From a more abstract perspective, the reason we can define notions such as completeness is because a topological vector space has something better than a topological structure, namely a uniform structure.)

Remark 1 As we have seen in previous lectures, complete normed vector spaces (i.e. Banach spaces) enjoy some very nice properties. Some of these properties (e.g. the uniform boundedness principle and the open mapping theorem extend to a slightly larger class of complete topological vector spaces, namely the Fréchet spaces. A Fréchet space is a complete Hausdorff topological vector space whose topology is generated by an at most countable family of semi-norms; examples include the space from Exercise 5 or the uniform convergence on compact sets topology from Exercise 4 in the case when is -compact. We will however not study Fréchet spaces systematically here.

One can also extend the notion of a dual space from normed vector spaces to topological vector spaces in the obvious manner: the dual space of a topological space is the space of continuous linear functionals from to the field of scalars (either or , depending on whether is a real or complex vector space). This is clearly a vector space. Unfortunately, in the absence of a norm on , one cannot define the analogue of the norm topology on ; but as we shall see below, there are some weaker topologies that one can still place on this dual space.

— 2. Compactness in the strong topology —

We now return to normed vector spaces, and briefly discuss compactness in the strong (or norm) topology on such spaces. In finite dimensions, the Heine-Borel theorem tells us that a set is compact if and only if it is closed and bounded. In infinite dimensions, this is not enough, for two reasons. Firstly, compact sets need to be complete, so we are only likely to find many compact sets when the ambient normed vector space is also complete (i.e. it is a Banach space). Secondly, compact sets need to be totally bounded, rather than merely bounded, and this is quite a stringent condition. Indeed it forces compact sets to be “almost finite-dimensional” in the following sense:

Exercise 11 Let be a subset of a Banach space . Show that the following are equivalent:

• (i) is compact.
• (ii) is sequentially compact.
• (iii) is closed and bounded, and for every , lies in the -neighbourhood of a finite-dimensional subspace of .

Suppose furthermore that there is a nested sequence of finite-dimensional subspaces of such that is dense. Show that the following statement is equivalent to the first three:

• (iv) is closed and bounded, and for every there exists an such that lies in the -neighbourhood of .

Example 6 Let . In order for a set to be compact in the strong topology, it needs to be closed and bounded, and also uniformly -power integrable at spatial infinity in the sense that for every there exists such that

for all . Thus, for instance, the “moving bump” example , where is the sequence which equals on and zero elsewhere, is not uniformly power integrable and thus not a compact subset of , despite being closed and bounded.

For “continuous” spaces, such as , uniform integrability at spatial infinity is not sufficient to force compactness in the strong topology; one also needs some uniform integrability at very fine scales, which can be described using harmonic analysis tools such as the Fourier transform. We will not discuss this topic here.

Exercise 12 Let be a normed vector space.

• If is a finite-dimensional subspace of , and , show that there exists such that for all . Give an example to show that is not necessarily unique (in contrast to the situation with Hilbert spaces).
• If is a finite-dimensional proper subspace of , show that there exists with such that for all . (cf. the Riesz lemma.)
• Show that the closed unit ball is compact in the strong topology if and only if is finite-dimensional.

— 3. The weak and weak* topologies —

Let be a topological vector space. Then, as discussed above, we have the vector space of continuous linear functionals on . We can use this dual space to create two useful topologies, the weak topology on and the weak* topology on :

Definition 2 (Weak and weak* topologies) Let be a topological vector space, and let be its dual.

• The weak topology on is the topology generated by the seminorms for all .
• The weak* topology on is the topology generated by the seminorms for all .

Remark 2 It is possible for two non-isomorphic topological vector spaces to have isomorphic duals, but with non-isomorphic weak* topologies. (For instance, has a very large number of preduals, which can generate a number of different weak* topologies on .) So, technically, one cannot talk about the weak* topology on a dual space , without specifying exactly what the predual space is. However, in practice, the predual space is usually clear from context.

Exercise 13 Show that the weak topology on is a topological vector space structure on that is weaker than the strong topology on . Also, if (and hence and ) are normed vector spaces, show that the weak* topology on is a topological vector space structure on that is weaker than the weak topology on (which is defined using the double dual . When is a reflexive normed vector space, show that the weak and weak* topologies on are equivalent.

From the definition, we see that a sequence converges in the weak topology, or converges weakly for short, to a limit if and only if for all . This weak convergence is often denoted , to distinguish it from strong convergence . Similarly, a sequence converges in the weak* topology to if for all (thus , viewed as a function on , converges pointwise to ).

Remark 3 If is a Hilbert space, then from the Riesz representation theorem for Hilbert spaces we see that a sequence converges weakly (or in the weak* sense) to a limit if and only if for all .

Exercise 14 Show that if is a normed vector space, then the weak topology on and the weak* topology on are both Hausdorff. (Hint: You will need the Hahn-Banach theorem.) In particular, we conclude the important fact that weak and weak* limits, when they exist, are unique.

The following exercise shows that the strong, weak, and weak* topologies can all differ from each other.

Exercise 15 Let , thus and . Let be the standard basis of either , , or .

• Show that the sequence converges weakly in to zero, but does not converge strongly in .
• Show that the sequence converges in the weak* sense in to zero, but does not converge in the weak or strong senses in .
• Show that the sequence for converges in the weak* topology of to zero, but does not converge in the weak or strong senses. (Hint: use a generalised limit functional).

Remark 4 Recall from Exercise 11 of Notes 9 that sequences in which converge in the weak topology, also converge in the strong topology. We caution however that the two topologies are not quite equivalent; for instance, the open unit ball in is open in the strong topology, but not in the weak.

Exercise 16 Let be a normed vector space, and let be a subset of . Show that the following are equivalent:

• is strongly bounded (i.e. is contained in a ball).
• is weakly bounded (i.e. is bounded for all ).

(Hint: use the Hahn-Banach theorem and the uniform boundedness principle.) Similarly, if is a subset of , and is a Banach space, show that is strongly bounded if and only if is weak* bounded (i.e. is bounded for each ).) Conclude in particular that any sequence which is weakly convergent in or weak* convergent in is necessarily bounded.

Exercise 17 Let be a Banach space, and let converge weakly to a limit . Show that the sequence is bounded, and

Observe from Exercise 15 that strict inequality can hold (cf. Fatou’s lemma). Similarly, if converges in the weak* topology to a limit , show that the sequence is bounded and that

Again, construct an example to show that strict inequality can hold. Thus we see that weak or weak* limits can lose mass in the limit, as opposed to strong limits (note from the triangle inequality that if converges strongly to , then converges to ).

Exercise 18 Let be a Hilbert space, and let converge weakly to a limit . Show that the following statements are equivalent:

• converges strongly to .
• converges to .

Exercise 19 Let be a separable Hilbert space. We say that a sequence converges in the Césaro sense to a limit if converges strongly to as .

• Show that if converges strongly to , then it also converges in the Césaro sense to .
• Give examples to show that weak convergence does not imply Césaro convergence, and vice versa. On the other hand, if a sequence converges both weakly and in the Césaro sense, show that the weak limit is necessarily equal to the Césaro limit.
• Show that a sequence converges weakly to if and only if every subsequence has a further subsequence that converges in the Césaro sense to .

Exercise 20 Let be a Banach space. Show that the closed unit ball in is also closed in the weak topology, and the closed unit ball in is closed in the weak* topology.

Exercise 21 Let be a Banach space. Show that the weak* topology on is complete.

Exercise 22 Let be a normed vector space, let be a subspace of which is closed in the strong topology of .

• Show that is closed in the weak topology of .
• If is a sequence and , show that converges to in the weak topology of if and only if it converges to in the weak topology of . (Because of this fact, we can often refer to “the weak topology” without specifying the ambient space precisely.)

Exercise 23 Let with the uniform (i.e. ) norm, and identify the dual space with in the usual manner.

• Show that a sequence converges weakly to a limit if and only if the are bounded in and converge pointwise to .
• Show that a sequence converges in the weak* topology to a limit if and only if the are bounded in and converge pointwise to .
• Show that the weak topology in is not complete.

(More generally, it may help to think of the weak and weak* topologies as being analogous to pointwise convergence topologies.)

One of the main reasons why we use the weak and weak* topologies in the first place is that they have much better compactness properties than the strong topology, thanks to the Banach-Alaoglu theorem:

Theorem 3 (Banach-Alaoglu theorem) Let be a normed vector space. Then the closed unit ball of is compact in the weak* topology.

This result should be contrasted with Exercise 12.

Proof: Let’s say is a complex vector space (the case of real vector spaces is of course analogous). Let be the closed unit ball of , then any linear functional maps the closed unit ball of into the disk . Thus one can identify with a subset of , the space of functions from to . One easily verifies that the weak* topology on is nothing more than the product topology of restricted to . Also, one easily shows that is closed in . But by Tychonoff’s theorem, is compact, and so is compact also.

One should caution that the Banach-Alaoglu theorem does not imply that the space is locally compact in the weak* topology, because the norm ball in has empty interior in the weak* topology unless is finite dimensional. In fact, we have the following result of Riesz:

Exercise 24 Let be a locally compact Hausdorff topological vector space. Show that is finite dimensional. (Hint: If is locally compact, then there exists an open neighbourhood of the origin whose closure is compact. Show that for some finite-dimensional subspace , where . Iterate this to conclude that for any . On the other hand, use the compactness of to show that for any point there exists such that is disjoint from . Conclude that and thence that .)

The sequential version of the Banach-Alaoglu theorem is also of importance (particularly in PDE):

Theorem 4 (Sequential Banach-Alaoglu theorem) Let be a separable normed vector space. Then the closed unit ball of is sequentially compact in the weak* topology.

Proof: The functionals in are uniformly bounded and uniformly equicontinuous on , which by hypothesis has a countable dense subset . By the sequential Tychonoff theorem, any sequence in then has a subsequence which converges pointwise on , and thus converges pointwise on by Exercise 28 of Notes 10, and thus converges in the weak* topology. But as is closed in this topology, we conclude that is sequentially compact as required.

Remark 5 One can also deduce the sequential Banach-Alaoglu theorem from the general Banach-Alaoglu theorem by observing that the weak* topology on the dual of a separable space is metrisable. The sequential Banach-Alaoglu theorem can break down for non-separable spaces. For instance, the closed unit ball in is not sequentially compact in the weak* topology, basically because the space of ultrafilters is not sequentially compact (see Exercise 12 of these lecture notes).

If is reflexive, then the weak topology on is identical to the weak* topology on . We thus have

Corollary 5 If is a reflexive normed vector space, then the closed unit ball in is weakly compact, and (if is separable) is also sequentially weakly compact.

Remark 6 If is a normed vector space that is not separable, then one can show that is not separable either. Indeed, using transfinite induction on first uncountable ordinal, one can construct an uncountable proper chain of closed separable subspaces of the inseparable space , which by the Hahn-Banach theorem induces an uncountable proper chain of closed subspaces on , which is not compatible with separability. As a consequence, a reflexive space is separable if and only if its dual is separable. [On the other hand, separable spaces can have non-separable duals; consider , for instance.]

In particular, any bounded sequence in a reflexive separable normed vector space has a weakly convergent subsequence. This fact leads to the very useful weak compactness method in PDE and calculus of variations, in which a solution to a PDE or variational problem is constructed by first constructing a bounded sequence of “near-solutions” or “near-extremisers” to the PDE or variational problem, and then extracting a weak limit. However, it is important to caution that weak compactness can fail for non-reflexive spaces; indeed, for such spaces the closed unit ball in may not even be weakly complete, let alone weakly compact, as already seen in Exercise 23. Thus, one should be cautious when applying the weak compactness method to a non-reflexive space such as or . (On the other hand, weak* compactness does not need reflexivity, and is thus safer to use in such cases.)

In later notes we will see that the (sequential) Banach-Alaoglu theorem will combine very nicely with the Riesz representation theorem for measures, leading in particular to Prokhorov’s theorem.

— 4. The strong and weak operator topologies —

Now we turn our attention from function spaces to spaces of operators. Recall that if and are normed vector spaces, then is the space of bounded linear transformations from to . This is a normed vector space with the operator norm

This norm induces the operator norm topology on . Unfortunately, this topology is so strong that it is difficult for a sequence of operators to converge to a limit; for this reason, we introduce two weaker topologies.

Definition 6 (Strong and weak operator topologies) Let be normed vector spaces. The strong operator topology on is the topology induced by the seminorms for all . The weak operator topology on is the topology induced by the seminorms for all and .

Note that a sequence converges in the strong operator topology to a limit if and only if strongly in for all , and converges in the weak operator topology. (In contrast, converges to in the operator norm topology if and only if converges to uniformly on bounded sets.) One easily sees that the weak operator topology is weaker than the strong operator topology, which in turn is (somewhat confusingly) weaker than the operator norm topology.

Example 7 When is the scalar field, then is canonically isomorphic to . In this case, the operator norm and strong operator topology coincide with the strong topology on , and the weak operator norm topology coincides with the weak topology on . Meanwhile, coincides with , and the operator norm topology coincides with the strong topology on , while the strong and weak operator topologies correspond with the weak* topology on .

We can rephrase the uniform boundedness principle for convergence (Corollary 1 from Notes 9) as follows:

Proposition 7 (Uniform boundedness principle) Let be a sequence of bounded linear operators from a Banach space to a normed vector space , let be another bounded linear operator, and let be a dense subspace of . Then the following are equivalent:

• converges in the strong operator topology of to .
• is bounded in the operator norm (i.e. is bounded), and the restriction of to converges in the strong operator topology of to the restriction of to .

Exercise 25 Let the hypotheses be as in Proposition 7, but now assume that is also a Banach space. Show that the conclusion of Proposition 7 continues to hold if “strong operator topology” is replaced by “weak operator topology”.

Exercise 26 Show that the operator norm topology, strong operator topology, and weak operator topology, are all Hausdorff. As these topologies are nested, we thus conclude that it is not possible for a sequence of operators to converge to one limit in one of these topologies and to converge to a different limit in another.

Example 8 Let , and for each , let be the translation operator by : . If is continuous and compactly supported, then (e.g. from dominated convergence) we see that in as . Since the space of continuous and compactly supported functions is dense in , this implies (from the above proposition, with some obvious modifications to deal with the continuous parameter instead of the discrete parameter ) that converges in the strong operator topology (and hence weak operator topology) to the identity. On the other hand, does not converge to the identity in the operator norm topology. Indeed, observe for any that , and thus .

In a similar vein, does not converge to anything in the strong operator topology (and hence does not converge in the operator norm topology either) in the limit , since (say) does not converge strongly in . However, one easily verifies that as for any compactly supported , and hence for all by the usual limiting argument, and hence converges in the weak operator topology to zero.

The following exercise may help clarify the relationship between the operator norm, strong operator, and weak operator topologies.

Exercise 27 Let be a Hilbert space, and let be a sequence of bounded linear operators.

• Show that in the operator norm topology if and only if for any bounded sequences .
• Show that in the strong operator topology if and only if for any convergent sequence and any bounded sequence .
• Show that in the weak operator topology if and only if for any convergent sequences .
• Show that in the operator norm (resp. weak operator) topology if and only if in the operator norm (resp. weak operator) topology. Give an example to show that the corresponding claim for the strong operator topology is false.

There is a counterpart of the Banach-Alaoglu theorem (and its sequential analogue), at least in the case of Hilbert spaces:

Exercise 28 Let be Hilbert spaces. Show that the closed unit ball (in the operator norm) in is compact in the weak operator topology. If and are separable, show that is sequentially compact in the weak operator topology.

The behaviour of convergence in various topologies with respect to composition is somewhat complicated, as the following exercise shows.

Exercise 29 Let be a Hilbert space, let be sequences of operators, and let be another operator.

• If in the operator norm (resp. strong operator or weak operator) topology, show that and in the operator norm (resp. strong operator or weak operator) topology.
• If in the operator norm topology, and is bounded in the operator norm topology, show that and in the operator norm topology.
• If in the strong operator topology, and is bounded in the operator norm topology, show that in the strong operator norm topology.
• Give an example where in the strong operator topology, and in the weak operator topology, but does not converge to zero even in the weak operator topology.

Exercise 30 Let be a Hilbert space. An operator is said to be finite rank if its image is finite dimensional. is said to be compact if the image of the unit ball is precompact. Let denote the space of compact operators on .

• Show that is compact if and only if it is the limit of finite rank operators in the operator norm topology. Conclude in particular that is a closed subset of in the operator norm topology.
• Show that an operator is compact if and only if is compact.
• If is separable, show that every is the limit of finite rank operators in the strong operator topology.
• If , show that maps weakly convergent sequences to strongly convergent sequences. (This property is known as complete continuity.)
• Show that is a subspace of , which is closed with respect to left and right multiplication by elements of . (In other words, the space of compact operators is an two-ideal in the algebra of bounded operators.)

The weak operator topology plays a particularly important role on the theory of von Neumann algebras, which we will not discuss here. We will return to the study of compact operators next quarter, when we discuss the spectral theorem.

[Update, Feb 23: Corrections, another exercise and remark added (note renumbering).]

Exercise 1 (Basic properties of compact sets)

• Show that any finite set is compact.
• Show that any finite union of compact subsets of a topological space is still compact.
• Show that any image of a compact space under a continuous map is still compact.

Show that these three statements continue to hold if “compact” is replaced by “sequentially compact”.

— 1. Compactness and the Hausdorff property —

Recall from Notes 8 that a topological space is Hausdorff if every distinct pair of points can be separated by two disjoint open neighbourhoods of respectively; every metric space is Hausdorff, but not every topological space is.

At first glance, the Hausdorff property bears no resemblance to the compactness property. However, they are in some sense “dual” to each other, as the following two exercises show:

Exercise 2 Let be a compact topological space.

• Show that every closed subset in is compact.
• Show that any weaker topology on also yields a compact topological space .
• Show that the trivial topology on is always compact.

Exercise 3 Let be a Hausdorff topological space.

• Show that every compact subset of is closed.
• Show that any stronger topology on also yields a Hausdorff topological space .
• Show that the discrete topology on is always Hausdorff.

The first exercise asserts that compact topologies tend to be weak, while the second exercise asserts that Hausdorff topologies tend to be strong. The next lemma asserts that the two concepts only barely overlap:

Lemma 1 Let be a weak and strong topology respectively on a space . If is compact and is Hausdorff, then . (In other words, a compact topology cannot be strictly stronger than a Hausdorff one, and a Hausdorff topology cannot be strictly weaker than a compact one.)

Proof: Since , every set which is closed in is closed in , and every set which is compact in is compact in . But from Exercises 2, 3, every set which is closed in is compact in , and every set which is compact in is closed in . Putting all this together, we see that and have exactly the same closed sets, and thus have exactly the same open sets; in other words, .

Corollary 2 Any continuous bijection from a compact topological space to a Hausdorff topological space is a homeomorphism.

Proof: Consider the pullback of the topology on by ; this is a topology on . As is continuous, this topology is weaker than , and thus by Lemma 1 is equal to . As is a bijection, this implies that is continuous, and the claim follows.

One may wish to compare this corollary with Corollary 2 from Notes 9.

Remark 1 Spaces which are both compact and Hausdorff (e.g. the unit interval with the usual topology) have many nice properties and are moderately common, so much so that the two properties are often concatenated as CH. Spaces that are locally compact and Hausdorff (e.g. manifolds) are much more common and have nearly as many nice properties, and so these two properties are often concatenated as LCH. One should caution that (somewhat confusingly) in some older literature (particularly those in the French tradition), “compact” is used for “compact Hausdorff”.

(Optional) Another way to contrast compactness and the Hausdorff property is via the machinery of ultrafilters. Define an filter on a space to be a collection of sets of which is closed under finite intersection, is also monotone (i.e. if and , then ), and does not contain the empty set. Define an ultrafilter to be a filter with the additional property that for any , exactly one of and lies in . (See also this blog post of mine for more discussion of ultrafilters.)

Exercise 4 (Ultrafilter lemma) Show that every filter is contained in at least one ultrafilter. (Hint: use Zorn’s lemma, see Notes 7.)

Exercise 5 A collection of subsets of has the finite intersection property if every finite intersection of sets in the collection has non-empty intersection. Show that every filter has the finite intersection property, and that every collection of sets with the finite intersection property is contained in a filter (and hence contained in an ultrafilter, by the ultrafilter lemma).

Given a point and an ultrafilter on , we say that converges to if every neighbourhood of belongs to .

Exercise 6 Show that a space is Hausdorff if and only if every ultrafilter has at most one limit. (Hint: For the “if” part, observe that if cannot be separated by disjoint neighbourhoods, then the neighbourhoods of and together enjoy the finite intersection property.)

Exercise 7 Show that a space is compact if and only if every ultrafilter has at least one limit. (Hint: use the finite intersection property formulation of compactness and Exercise 5.)

— 2. Compactness and bases —

Compactness is the property that every open cover has a finite subcover. This property can be difficult to verify in practice, in part because the class of open sets is very large. However, in many cases one can replace the class of open sets with a much smaller class of sets. For instance, in metric spaces, a set is open if and only if it is the union of open balls (note that the union may be infinite or even uncountable). We can generalise this notion as follows:

Definition 3 (Base) Let be a topological space. A base for this space is a collection of open sets such that every open set in can be expressed as the union of sets in the base. The elements of are referred to as basic open sets.

Example 1 The collection of open balls in a metric space forms a base for the topology of that space. As another (rather trivial) example of a base: any topology is a base for itself.

This concept should be compared with that of a basis of a vector space: every vector in that space can be expressed as a linear combination of vectors in a basis. However, one difference between a base and a basis is that the representation of an open set as the union of basic open sets is almost certainly not unique.

Given a base , define a basic open neighbourhood of a point to be a basic open set that contains . Observe that a set is open if and only if every point in has a basic open neighbourhood contained in .

Exercise 8 Let be a collection of subsets of a set . Show that is a basis for some topology if and only if it it covers and has the following additional property: given any and any two basic open neighbourhoods of , there exists another basic open neighbourhood of that is contained in . Furthermore, the topology is uniquely determined by .

To verify the compactness property, it suffices to do so for basic open covers (i.e. coverings of the whole space by basic open sets):

Exercise 9 Let be a topological space with a base . Then the following are equivalent:

• Every open cover has a finite subcover (i.e. is compact);
• Every basic open cover has a finite subcover.

A useful fact about compact metric spaces is that they are in some sense “countably generated”.

Lemma 4 Let be a compact metric space.

• (i) is separable (i.e. it has an at most countably infinite dense subset).
• (ii) is second-countable (i.e. it has an at most countably infinite base).

Proof: By Theorem 1 of Notes 8, is totally bounded. In particular, for every , one can cover by a finite number of balls of radius . The set of points is then easily verified to be dense and at most countable, giving (i). Similarly, the set of balls can be easily verified to be a base which is at most countable, giving (ii).

Remark 2 One can easily generalise compactness here to -compactness; thus for instance finite-dimensional vector spaces are separable and second-countable. The properties of separability and second-countability are much weaker than -compactness in general, but can still serve to provide some constraint as to the “size” or “complexity” of a metric space or topological space in many situations.

We now weaken the notion of a base to that of a sub-base.

Definition 5 (Sub-base) Let be a topological space. A sub-base for this space is a collection of subsets of such that is the weakest topology that makes open (i.e. is generated by ). Elements of are referred to as sub-basic open sets.

Observe for instance that every base is a sub-base. The converse is not true: for instance, the half-open intervals for form a sub-base for the standard topology on , but not a base. In contrast to bases, which need to obey the property in Exercise 8, no property is required on a collection in order for it to be a sub-base; every collection of sets generates a unique topology with respect to which it is a sub-base.

The precise relationship between sub-bases and bases is given by the following exercise.

Exercise 10 Let be a topological space, and let be a collection of subsets of . Then the following are equivalent:

• is a sub-base for .
• The space of finite intersections of (including the whole space , which corresponds to the case ) is a base for .

Thus a set is open iff it is the union of finite intersections of sub-basic open sets.

Many topological facts involving open sets can often be reduced to verifications on basic or sub-basic open sets, as the following exercise illustrates:

Exercise 11 Let be a topological space, and be a sub-base of , and let be a base of .

• Show that a sequence converges to a limit if and only if every sub-basic open neighbourhood of contains for all sufficiently large . (Optional: show that an analogous statement is also true for nets.)
• Show that a point is adherent to a set if and only if every basic open neighbourhood of intersects . Give an example to show that the claim fails for sub-basic open sets.
• Show that a point is in the interior of a set if and only if contains a basic open neighbourhood of . Give an example to show that the claim fails for sub-basic open sets.
• If is another topological space, show that a map is continuous if and only if the inverse image of every sub-basic open set is open.

There is a useful strengthening of Exercise 9 in the spirit of the above exercise, namely the Alexander sub-base theorem:

Theorem 6 (Alexander sub-base theorem) Let be a topological space with a sub-base . Then the following are equivalent:

• Every open cover has a finite subcover (i.e. is compact);
• Every sub-basic open cover has a finite subcover.

Proof: Call an open cover bad if it had no finite subcover, and good otherwise. In view of Exercise 9, it suffices to show that if every sub-basic open cover is good, then every basic open cover is good also, where we use the basis coming from Exercise 10.

Suppose for contradiction that every sub-basic open cover was good, but at least one basic open cover was bad. If we order the bad basic open covers by set inclusion, observe that every chain of bad basic open covers has an upper bound that is also a bad basic open cover, namely the union of all the covers in the chain. Thus, by Zorn’s lemma, there exists a maximal bad basic open cover . Thus this cover has no finite subcover, but if one adds any new basic open set to this cover, then there must now be a finite subcover.

Pick a basic open set in this cover . Then we can write for some sub-basic open sets . We claim that at least one of the also lie in the cover . To see this, suppose for contradiction that none of the was in . Then adding any of the to enlarges the basic open cover and thus creates a finite subcover; thus together with finitely many sets from cover , or equivalently that one can cover with finitely many sets from . Thus one can also cover with finitely many sets from , and thus itself can be covered by finitely many sets from , a contradiction.

From the above discussion and the axiom of choice, we see that for each basic set in there exists a sub-basic set containing that also lies in . (Two different basic sets could lead to the same sub-basic set , but this will not concern us.) Since the cover , the do also. By hypothesis, a finite number of can cover , and so is good, which gives the desired a contradiction.

Exercise 12 (Optional) Use Exercise 7 to give another proof of the Alexander sub-base theorem.

Exercise 13 Use the Alexander sub-base theorem to show that the unit interval (with the usual topology) is compact, without recourse to the Heine-Borel or Bolzano-Weierstrass theorems.

Exercise 14 Let be a well-ordered set, endowed with the order topology (Exercise 9 of Notes 8); such a space is known as an ordinal space. Show that is Hausdorff, and that is compact if and only if has a maximal element.

One of the major applications of the sub-base theorem is to prove Tychonoff’s theorem, which we turn to next.

— 3. Compactness and product spaces —

Given two topological spaces and , we can form the product space , using the cylinder sets as a sub-base, or equivalently using the open boxes as a base (cf. Example 15 from Notes 8). One easily verifies that the obvious projection maps , are continuous, and that these maps also provide homeomorphisms between and , or between and , for every . Also observe that a sequence (or net ) converges to a limit in if and only if and (or and ) converge in and to and respectively.

This operation preserves a number of useful topological properties, for instance

Exercise 15 Prove that the product of two Hausdorff spaces is still Hausdorff.

Exercise 16 Prove that the product of two sequentially compact spaces is still sequentially compact.

Proposition 7 The product of two compact spaces is compact.

Proof: By Exercise 9 it suffices to show that any basic open cover of by boxes has a finite subcover. For any , this open cover covers ; by the compactness of , we can thus cover by a finite number of open boxes . Intersecting the together, we obtain a neighbourhood of such that is covered by a finite number of these boxes. But by compactness of , we can cover by a finite number of . Thus all of can be covered by a finite number of boxes in the cover, and the claim follows.

Exercise 17 (Optional) Obtain an alternate proof of this proposition using Exercise 14 from Notes 8.

The above theory for products of two spaces extends without difficulty to products of finitely many spaces. Now we consider infinite products.

Definition 8 (Product spaces) Given a family of topological spaces, let be the Cartesian product, i.e. the space of tuples with for all . For each , we have the obvious projection map that maps to .

• We define the product topology on to be the topology generated by the cylinder sets for and as a sub-base, or equivalently the weakest topology that makes all of the continuous.
• We define the box topology on to be the topology generated by all the boxes , where for all .

Unless otherwise specified, we assume the product space to be endowed with the product topology rather than the box topology.

When is finite, the product topology and the box topology coincide. When is infinite, the two topologies are usually different (as we shall see), but the box topology is always at least as strong as the product topology. Actually, in practice the box topology is too strong to be of much use – there are not enough convergent sequences in it. For instance, in the space of real-valued sequences , even sequences such as do not converge to the zero sequence as (why?), despite converging in just about every other sense.

Exercise 18 Show that the arbitrary product of Hausdorff spaces remains Hausdorff in either the product or the box topology.

Exercise 19 Let be a sequence of metric spaces. Show that the the function on the product space defined by

is a metric on which generates the product topology on .

Exercise 20 Let be a product space with the product topology. Show that a sequence in that space converges to a limit if and only if converges in to for every . (The same statement also holds for nets.) Thus convergence in the product topology is essentially the same concept as pointwise convergence (cf. Example 14 of Notes 8).

The box topology usually does not preserve compactness. For instance, one easily checks that the product of any number of discrete spaces is still discrete in the box topology. On the other hand, a discrete space is compact (or sequentially compact) if and only if it is finite. Thus the infinite product of any number of non-trivial (i.e. having at least two elements) compact discrete spaces will be non-compact, and similarly for sequential compactness.

The situation improves significantly with the product topology, however (which is weaker, and thus more likely to be compact). We begin with the situation for sequential compactness.

Proposition 9 (Sequential Tychonoff theorem) Any at most countable product of sequentially compact topological spaces is sequentially compact.

Proof: We will use the “Arzelá-Ascoli diagonalisation argument”. The finite case is already handled by Exercise 16 (and can in any event be easily deduced from the countable case), so suppose we have a countably infinite sequence of sequentially compact spaces, and consider the product space with the product topology. Let be a sequence in , thus each is itself a sequence with for all . Our objective is to find a subsequence which converges to some limit in the product topology, which by Exercise 20 is the same as pointwise convergence (i.e. as for each ).

Consider the first coordinates of the sequence . As is sequentially compact, we can find a subsequence in such that converges in to some limit .

Now, in this subsequence, consider the second coordinates . As is sequentially compact, we can find a further subsequence in such that converges in to some limit . Also, we inherit from the preceding subsequence that converges in to .

We continue in this vein, creating nested subsequences for whose first components converge to respectively.

None of these subsequences, by themselves are sufficient to finish the problem. But now we use the diagonalisation trick: we consider the diagonal sequence . One easily verifies that converges in to as for every , and so we have extracted a sequence that is convergent in the product topology.

Remark 3 In the converse direction, if a product of spaces is sequentially compact, then each of the factor spaces must also be sequentially compact, since they are continuous images of the product space and one can apply Exercise 1.

The sequential Tychonoff theorem breaks down for uncountable products. Consider for instance the product space of functions . As (with the discrete topology) is sequentially compact, this is an (uncountable) product of sequentially compact spaces. On the other hand, for each we can define the evaluation function by . This is a sequence in ; we claim that it has no convergent subsequence. Indeed, given any , we can find such that does not converge to a limit as , and so does not converge pointwise (i.e. does not converge in the product topology).

However, we can recover the result for uncountable products as long as we work with topological compactness rather than sequential compactness, leading to Tychonoff’s theorem:

Theorem 10 (Tychonoff theorem) Any product of compact topological spaces is compact.

Proof: Write for this product of compact topological spaces. By Theorem 6, it suffices to show that any open cover of by sub-basic open sets has a finite sub-cover, where is some index set, and for each , and is open in .

For each , consider the sub-basic open sets that are associated to those with . If the open sets here cover , then by compactness of , a finite number of the already suffice to cover , and so a finite number of the cover , and we are done. So we may assume that the do not cover , thus there exists that avoids all the with . One then sees that the point in avoids all of the , a contradiction. The claim follows.

Remark 4 The axiom of choice was used in several places in the proof (in particular, via the Alexander sub-base theorem). This turns out to be necessary, because one can use Tychonoff’s theorem to establish the axiom of choice. This was first observed by Kelley, and can be sketched as follows. It suffices to show that the product of non-empty sets is again non-empty. We can make each compact (e.g. by using the trivial topology). We then adjoin an isolated element to each to obtain another compact space , with closed in . By Tychonoff’s theorem, the product is compact, and thus every collection of closed sets with finite intersection property has non-empty intersection. But observe that the sets in , where is the obvious projection, are closed and has the finite intersection property; thus the intersection of all of these sets is non-empty, and the claim follows.

Remark 5 From the above discussion, we see that the space is compact but not sequentially compact; thus compactness does not necessarily imply sequential compactness.

Exercise 21 Let us call a topological space first-countable if, for every , there exists a countable family of open neighbourhoods of such that every neighbourhood of contains at least one of the .

• Show that every metric space is first-countable.
• Show that every second-countable space is first-countable (see Lemma 4).
• Show that every separable metric space is second-countable.
• Show that every space which is second-countable, is separable.
• (Optional) Show that every net which converges in to , has a convergent subsequence (i.e. a subnet whose index set is ).
• Show that any compact space which is first-countable, is also sequentially compact. (The converse is not true: Exercise 9 from Notes 8 provides a counterexample.)

(Optional) There is an alternate proof of the Tychonoff theorem that uses the machinery of universal nets. We sketch this approach in a series of exercises.

Definition 11 A net in a set is universal if for every function , the net converges to either or .

Exercise 22 Show that a universal net in a compact topological space is necessarily convergent. (Hint: show that the collection of closed sets which contain for sufficiently large enjoys the finite intersection property.)

Exercise 23 (Kelley’s theorem) Every net in a set has a universal subnet . (Hint: First use Exercise 5 to find an ultrafilter on that contains the upsets for all . Now let be the space of all pairs , where , ordered by requiring when and , and let be the map .)

Exercise 24 Use the previous two exercises, together with Exercise 20, to establish an alternate proof of Tychonoff’s theorem.

Exercise 25 Establish yet another proof of Tychonoff’s theorem using Exercise 7 directly (rather than proceeding via Exercise 12).

— 4. Compactness and equicontinuity —

We now pause to give an important application of the (sequential) Tychonoff theorem. We begin with some definitions. If is a topological space and is a metric space, let be the space of bounded continuous functions from to . (If is compact, this is the same space as , the space of continuous functions from to .) We can give this space the uniform metric

Exercise 26 If is complete, show that is a complete metric space. (Note that this implies Exercise 2 from Notes 6.)

Note that if is continuous if and only if, for every and , there exists a neighbourhood of such that for all . We now generalise this concept to families.

Definition 12 Let be a topological space, let be a metric space, and Let be a family of functions .

• We say that this family is pointwise bounded if for every , the set is bounded in .
• We say that this family is pointwise precompact if for every , the set is precompact in .
• We say that this family is equicontinuous if for every and , there exists a neighbourhood of such that for all and .
• If is also a metric space, we say that the family is uniformly equicontinuous if for every there exists a such that for all and with .

Remark 6 From the Heine-Borel theorem, the pointwise boundedness and pointwise precompactness properties are equivalent if is a subset of for some . Any finite collection of continuous functions is automatically an equicontinuous family (why?), and any finite collection of uniformly continuous functions is automatically a uniformly equicontinuous family; the concept only acquires additional meaning once one considers infinite families of continuous functions.

Example 2 With and , the family of functions for are pointwise bounded (and thus pointwise precompact), but not equicontinuous. The family of functions for , on the other hand, are equicontinuous, but not pointwise bounded or pointwise precompact. The family of functions for are pointwise bounded (even uniformly bounded), but not equicontinuous.

Example 3 With and , the functions are pointwise bounded (even uniformly bounded), are equicontinuous, and are each individually uniformly continuous, but are not uniformly equicontinuous.

Exercise 27 Show that the uniform boundedness principle (Theorem 2 from Notes 9) can be restated as the assertion that any family of bounded linear operators from the unit ball of a Banach space to a normed vector space is pointwise bounded if and only if it is equicontinuous.

Example 4 A function between two metric spaces is said to be Lipschitz (or Lipschitz continuous) if there exists a constant such that for all ; the smallest constant one can take here is known as the Lipschitz constant of . Observe that Lipschitz functions are automatically continuous, hence the name. Also observe that a family of Lipschitz functions with uniformly bounded Lipschitz constant is equicontinuous.

One nice consequence of equicontinuity is that it equates uniform convergence with pointwise convergence, or even pointwise convergence on a dense subset.

Exercise 28 Let be a topological space, let be a complete metric space, let be an equicontinuous family of functions. Show that the following are equivalent:

• The sequence is pointwise convergent.
• The sequence is pointwise convergent on some dense subset of .

If is compact, show that the above two statements are also equivalent to

• The sequence is uniformly convergent.

(Compare with Corollary 1 from Notes 9.) Show that no two of the three statements remain equivalent if the hypothesis of equicontinuity is dropped.

We can now use Proposition 9 to give a useful characterisation of precompactness in when is compact, known as the Arzelá-Ascoli theorem:

Theorem 13 (Arzelá-Ascoli theorem) Let be a metric space, be a compact metric space, and let be a family of functions . Then the following are equivalent:

• (i) is a precompact subset of .
• (ii) is pointwise precompact and equicontinuous.
• (iii) is pointwise precompact and uniformly equicontinuous.

Proof: We first show that (i) implies (ii). For any , the evaluation map is a continuous map from to , and thus maps precompact sets to precompact sets. As a consequence, any precompact family in is pointwise precompact. To show equicontinuity, suppose for contradiction that equicontinuity failed at some point , thus there exists , a sequence , and points such that for every . One then verifies that no subsequence of can converge uniformly to a continuous limit, contradicting precompactness. (Note that in the metric space , precompactness is equivalent to sequential precompactness.)

Now we show that (ii) implies (iii). It suffices to show that equicontinuity implies uniform equicontinuity. This is a straightforward generalisation of the more familiar argument that continuity implies uniform continuity on a compact domain, and we repeat it here. Namely, fix . For every , equicontinuity provides a such that whenever and . The balls cover , thus by compactness some finite subcollection , of these balls cover . One then easily verifies that whenever with .

Finally, we show that (iii) implies (i). It suffices to show that any sequence , , which is pointwise precompact and uniformly equicontinuous, has a convergent subsequence. By embedding in its metric completion , we may assume without loss of generality that is complete. (Note that for every , the set is precompact in , hence the closure in is complete and thus closed in also. Thus any pointwise limit of the in will take values in .) By Lemma 4, we can find a countable dense subset of . For each , we can use pointwise precompactness to find a compact set such that takes values in . For each , the tuple can then be viewed as a point in the product space . By Proposition 9, this product space is sequentially compact, hence we may find a subsequence such that is convergent in the product topology, or equivalently that pointwise converges on the countable dense set . The claim now follows from Exercise 28.

Remark 7 The above theorem characterises precompact subsets of when is a compact metric space. One can also characterise compact subsets by observing that a subset of a metric space is compact if and only if it is both precompact and closed.

There are many variants of the Arzelá-Ascoli theorem with stronger or weaker hypotheses or conclusions; for instance, we have

Corollary 14 (Arzelá-Ascoli theorem, special case) Let be a sequence of functions from a compact metric space to a finite-dimensional vector space which are equicontinuous and pointwise bounded. Then there is a subsequence of which converges uniformly to a limit (which is necessarily bounded and continuous).

Thus, for instance, any sequence of uniformly bounded and uniformly Lipschitz functions will have a uniformly convergent subsequence. This claim fails without the uniform Lipschitz assumption (consider, for instance, the functions ). Thus one needs a “little bit extra” uniform regularity in addition to uniform boundedness in order to force the existence of uniformly convergent subsequences. This is a general phenomenon in infinite-dimensional function spaces: compactness in a strong topology tends to require some sort of uniform control on regularity or decay in addition to uniform bounds on the norm.

Exercise 29 Show that the equivalence of (i) and (ii) continues to hold if is assumed to be just a compact Hausdorff space rather than a compact metric space (the statement (iii) no longer makes sense in this setting). Hint: need not be separable any more, however one can still adapt the diagonalisation argument used to prove Proposition 9. The starting point is the observation that for every and every , one can find a neighbourhood of and some subsequence which only oscillates by at most (or maybe ) on .

Exercise 30 (Locally compact Hausdorff version of Arzelá-Ascoli) Let be a locally compact Hausdorff space which is also -compact, and let be an equicontinuous, pointwise bounded sequence of functions. Then there exists a subsequence which converges uniformly on compact subsets of to a limit . (Hint: Express as a countable union of compact sets , each one contained in the interior of the next. Apply the compact Hausdorff Arzelá-Ascoli theorem on each compact set (Exercise 29). Then apply the Arzelá-Ascoli argument one last time.)

Remark 8 The Arzelá-Ascoli theorem (and other compactness theorems of this type) are often used in partial differential equations, to demonstrate existence of solutions to various equations or variational problems. For instance, one may wish to solve some equation , for some function . One way to do this is to first construct a sequence of approximate solutions, so that as in some suitable sense. If one can also arrange these to be equicontinuous and pointwise bounded, then the Arzelá-Ascoli theorem allows one to pass to a subsequence that converges to a limit . Given enough continuity (or semi-continuity) properties on , one can then show that as required.

More generally, the use of compactness theorems to demonstrate existence of solutions in PDE is known as the compactness method. It is applicable in a remarkably broad range of PDE problems, but often has the drawback that it is difficult to establish uniqueness of the solutions created by this method (compactness guarantees existence of a limit point, but not uniqueness). Also, in many cases one can only hope for compactness in rather weak topologies, and as a consequence it is often difficult to establish regularity of the solutions obtained via compactness methods.

[Update, Feb 20: some corrections, new exercise added (note renumbering).]

Lemma 1. (Pigeonhole principle for measure spaces) Let be an at most countable sequence of measurable subsets of a measure space X.  If has positive measure, then at least one of the has positive measure.

Now suppose that X was a Euclidean space with Lebesgue measure m.  The Lebesgue differentiation theorem easily implies that having positive measure is equivalent to being “dense” in certain balls:

Proposition 1. Let be a measurable subset of .  Then the following are equivalent:

1. E has positive measure.
2. For any , there exists a ball B such that .

Thus one can think of a null set as a set which is “nowhere dense” in some measure-theoretic sense.

It turns out that there are analogues of these results when the measure space   is replaced instead by a complete metric space .  Here, the appropriate notion of a “small” set is not a null set, but rather that of a nowhere dense set: a set E which is not dense in any ball, or equivalently a set whose closure has empty interior.  (A good example of a nowhere dense set would be a proper subspace, or smooth submanifold, of , or a Cantor set; on the other hand, the rationals are a dense subset of and thus clearly not nowhere dense.)   We then have the following important result:

Theorem 1. (Baire category theorem). Let be an at most countable sequence of subsets of a complete metric space X.  If contains a ball B, then at least one of the is dense in a sub-ball B’ of B (and in particular is not nowhere dense).  To put it in the contrapositive: the countable union of nowhere dense sets cannot contain a ball.

Exercise 1. Show that the Baire category theorem is equivalent to the claim that in a complete metric space, the countable intersection of open dense sets remain dense. 

Exercise 2. Using the Baire category theorem, show that any non-empty complete metric space without isolated points is uncountable.  (In particular, this shows that Baire category theorem can fail for incomplete metric spaces such as the rationals .) 

To quickly illustrate an application of the Baire category theorem, observe that it implies that one cannot cover a finite-dimensional real or complex vector space by a countable number of proper subspaces.  One can of course also establish this fact by using Lebesgue measure on this space.  However, the advantage of the Baire category approach is that it also works well in infinite dimensional complete normed vector spaces, i.e. Banach spaces, whereas the measure-theoretic approach runs into significant difficulties in infinite dimensions.  This leads to three fundamental equivalences between the qualitative theory of continuous linear operators on Banach spaces (e.g. finiteness, surjectivity, etc.) to the quantitative theory (i.e. estimates):

1. The uniform boundedness principle, that equates the qualitative boundedness (or convergence) of a family of continuous operators with their quantitative boundedness.
2. The open mapping theorem, that equates the qualitative solvability of a linear problem Lu = f with the quantitative solvability.
3. The closed graph theorem, that equates the qualitative regularity of a (weakly continuous) operator T with the quantitative regularity of that operator.

Strictly speaking, these theorems are not used much directly in practice, because one usually works in the reverse direction (i.e. first proving quantitative bounds, and then deriving qualitative corollaries); but the above three theorems help explain why we usually approach qualitative problems in functional analysis via their quantitative counterparts.

– Proof of Baire category theorem –

Assume that the Baire category theorem failed; then it would be possible to cover a ball in a complete metric space by a countable family of nowhere dense sets.

We now invoke the following easy observation: if E is nowhere dense, then every ball B contains a subball B’ which is disjoint from E.  Indeed, this follows immediately from the definition of a nowhere dense set.

Invoking this observation, we can find a ball in (say) which is disjoint from ; we may also assume that by shrinking as necessary.  Then, inside , we can find a ball which is also disjoint from , with .  Continuing this process, we end up with a nested sequence of balls , each of which are disjoint from , and such that and for all .

From the triangle inequality we have , and so the sequence is a Cauchy sequence.  As X is complete, converges to a limit x.  Summing the geometric series, one verifies that for all , and in particular is an element of B which avoids all of , a contradiction. 

We can illustrate the analogy between the Baire category theorem and the measure-theoretic analogs by introducing some further definitions.  Call a set E meager or of the first category if it can be expressed (or covered) by a countable union of nowhere dense sets, and of the second category if it is not meager.  Thus, the Baire category theorem shows that any subset of a complete metric space with non-empty interior is of the second category, which may help explain the name for the property.  Call a set co-meager or residual if its complement is meager, and call a set Baire or almost open if it differs from an open set by a meager set (note that a Baire set is unrelated to the Baire -algebra).  Then we have the following analogy between complete metric space topology, and measure theory:

Nowhere dense sets are meager, and meager sets have empty interior. Contrapositively, sets with dense interior
are residual, and residual sets are somewhere dense.  Taking complements instead of contrapositives, we see that open dense sets are co-meager,and co-meager sets are dense.

While there are certainly many analogies between meager sets and null sets (for instance, both classes are closed under countable unions, or under intersections with arbitrary sets), the two concepts can differ in practice.  For instance, in the real line with the standard metric and measure space structures, the set

(1)

where is an enumeration of the rationals, is open and dense, but has Lebesgue measure at most 2; thus its complement has infinite measure in but is nowhere dense (hence meager).  As a variant of this, the set

(2)

is a null set, but is the intersection of countably many open dense sets and is thus co-meager.

Exercise 3. A real number x is Diophantine if for every there exists such that for every rational number .  Show that  the set of Diophantine real numbers has full measure but is meager. 

Remark 1. If one assumes some additional axioms of set theory (e.g. the continuum hypothesis), it is possible to show that the collection of meager subsets of and the collection of null subsets of (viewed as -ideals of the collection of all subsets of ) are isomorphic; this is the Sierpinski-Erdös theorem, which we will not prove here.  Roughly speaking, this theorem tells us that any “effective” first-order statement which is true about meager sets will also be true about null sets, and conversely.

– The uniform boundedness principle –

As mentioned in the introduction, the Baire category theorem implies various equivalences between qualitative and quantitative properties of linear transformations between Banach spaces.  (Lemma 1 of Notes 3 already gives a prototypical such equivalence between a qualitative property (continuity) and a quantitative one (boundedness).)

Theorem 2. (Uniform boundedness principle)  Let X be a Banach space, let Y be a normed vector space, and let be a family of continuous linear operators .  Then the following are equivalent:

1. (Pointwise boundedness) For every , the set is bounded.
2. (Uniform boundedness) The operator norms are bounded.

The uniform boundedness principle is also known as the Banach-Steinhaus theorem.

Proof. It is clear that 2. implies 1.; now assume 1 holds and let us obtain 2.

For each , let be the set

. (3)

The hypothesis 1 is nothing more than the assertion that the cover X, and thus by the Baire category theorem must be dense in a ball.  Since the are continuous, the are closed, and so one of the contains a ball.  Since , we see that one of the contains a ball centred at the origin.  Dilating n as necessary, we see that one of the contains the unit ball .  But then all the are bounded by n, and the claim follows.

Exercise 4. Give counterexamples to show that the uniform boundedness principle fails if one relaxes the assumptions in any of the following ways:

1. X is merely a normed vector space rather than a Banach space (i.e. completeness is dropped).
2. The are not assumed to be continuous.
3. The are allowed to be nonlinear rather than linear.

Thus completeness, continuity, and linearity are all essential for the uniform boundedness principle to apply.

Remark 2. It is instructive to establish the uniform boundedness principle more “constructively” without the Baire category theorem (though the proof of the Baire category theorem is still implicitly present), as follows.  Suppose that 2 fails, then is unbounded.  We can then find a sequence such that (say) for all n.  We can then find unit vectors such that .

We can then form the absolutely convergent (and hence conditionally convergent, by completeness) sum for some choice of signs recursively as follows: once have been chosen, choose the sign so that

.  (4)

From the triangle inequality we soon conclude that

(5)

But by hypothesis, the RHS is unbounded in n, contradicting 1. 

A common way to apply the uniform boundedness principle is via the following corollary:

Corollary 1. (Uniform boundedness principle for norm convergence)  Let X be a Banach space, let Y be a normed vector space, and let be a family of continuous linear operators .  Then the following are equivalent:

1. (Pointwise convergence) For every , converges strongly in Y as .
2. (Pointwise convergence to a continuous limit) There exists a continuous linear such that for every , converges strongly in Y to Tx as .
3. (Uniform boundedness + dense subclass convergence) The operator norms are bounded, and for a dense set of x in X, converges strongly in Y as .

Proof. Clearly 2. implies 1., and as convergent sequences are bounded, we see from Theorem 1 that 1. implies 3.  The implication of 2 from 3 follows by a standard limiting argument and is left as an exercise. 

Remark 3. The same equivalences hold if one replaces the sequence by a net .

Example 1 (Fourier inversion formula).  For any and N > 0, define the Dirichlet summation operator

(4)

where is the Fourier transform of f, defined on smooth compactly supported functions by the formula and then extended to by the  Plancherel theorem.  Using the Plancherel identity, we can verify that the operator norms are uniformly bounded (indeed, they are all 1); also, one can check that for , that converges in norm to f as .  As is known to be dense in , this implies that converges in norm to f for every .

This argument only used the “easy” implication of Corollary 1, namely the deduction of 2. from 3.  The “hard” implication using the Baire category theorem was not directly utilised.  However, from a metamathematical standpoint, that implication is important because it tells us that the above strategy to prove convergence in norm of the Fourier inversion formula on – i.e. to obtain uniform operator norms on the partial sums, and to establish convergence on a dense subclass of “nice” functions – is in some sense the only strategy available to prove such a result. 

Remark 4. There is a partial analogue of Corollary 1 for the question of pointwise almost everywhere convergence rather than norm convergence, known as Stein’s maximal principle (discussed for instance in this previous blog post of mine).  For instance, it reduces Carleson’s theorem on the pointwise almost everywhere convergence of Fourier series to the boundedness of a certain maximal function (the Carleson maximal operator) related to Fourier summation, although the latter task is again quite non-trivial.  (As in Example 1, the role of the maximal principle is meta-mathematical rather than direct.)

Of course, if we omit some of the hypotheses, it is no longer true that pointwise boundedness and uniform boundedness are the same.  For instance, if we let be the space of complex sequences with only finitely many non-zero entries and with the uniform topology, and let be the map , then the are pointwise bounded but not uniformly bounded; thus completeness of X is important.  Also, even in the one-dimensional case , the uniform boundedness principle can easily be seen to fail if the are non-linear transformations rather than linear ones.

– The open mapping theorem –

A map between topological spaces X and Y is said to be open if it maps open sets to open sets.  This is similar to, but slightly different, from the more familiar property of being continuous, which is equivalent to the inverse image of open sets being open.  For instance, the map defined by is continuous but not open; conversely, the function defined by is discontinuous but open.

We have seen that it is quite possible for non-linear continuous maps to fail to be open.  But for linear maps between Banach spaces, the situation is much better:

Theorem 3. (Open mapping theorem)  Let be a continuous linear transformation between two Banach spaces X and Y.  Then the following are equivalent:

1. L is surjective.
2. L is open.
3. (Qualitative solvability) For every there exists a solution to the equation .
4. (Quantitative solvability) There exists a constant such that for every there exists a solution to the equation , which obeys the bound .
5. (Quantitative solvability for a dense subclass) There exists a constant such that for a dense set of f in Y, there exists a solution to the equation , which obeys the bound .

Proof. Clearly 4. implies 3., which is equivalent to 1., and it is easy to see from linearity that 2. and 4. are equivalent (cf. the proof of Lemma 1 from Notes 3).  4. trivially implies 5., while to obtain 4. from 5., observe that if E is any dense subset of the Banach space Y, then any f in Y can be expressed as an absolutely convergent series of elements in E (since one can iteratively approximate the residual to arbitrary accuracy by an element of E for ), and the claim easily follows.  So it suffices to show that 3. implies 4.

For each n, let be the set of all for which there exists a solution to Lu=f with .  From the hypothesis 3, we see that .  Since Y is complete, the Baire category theorem implies that there is some which is dense in some ball in Y.  In other words, the problem Lu=f is approximately quantitatively solvable in the ball in the sense that

• For every and every , there exists an approximate solution u with and , and thus .

By subtracting two such approximate solutions, we conclude that

• For any and any , there exists with and .

Since L is homogeneous, we can rescale and conclude that

• For any and any there exists with and .

In particular, setting (treating the case f=0 separately), we conclude that

• For any , we may write , where and .

We can iterate this procedure and then take limits (now using the completeness of X rather than Y) to obtain a solution to Lu=f for every with , and the claim follows.

Remark 5. The open mapping theorem provides metamathematical justification for the method of a priori estimates for solving linear equations such as for a given datum and for an unknown , which is of course a familiar problem in linear PDE.  The a priori method assumes that f is in some dense class of nice functions (e.g. smooth functions) in which solvability of Lu=f is presumably easy, and then proceeds to obtain the a priori estimate for some constant C.  Theorem 3 then assures that Lu=f is solvable for all f in Y (with a similar bound).  As before, this implication does not directly use the Baire category theorem, but that theorem helps explain why this method is “not wasteful”. 

A pleasant corollary of the open mapping theorem is that, as with ordinary linear algebra or with arbitrary functions, invertibility is the same thing as bijectivity:

Corollary 2. Let be a continuous linear operator between two Banach spaces X, Y.  Then the following are equivalent:

1. (Qualitative invertibility) T is bijective.
2. (Quantitative invertibility) T is bijective, and is a continuous (hence bounded) linear transformation.

Remark 6. The claim fails without the completeness hypotheses on X and Y.  For instance, consider the operator defined by , where we give the uniform norm.  Then T is continuous and bijective, but is unbounded.

Exercise 5. Show that Corollary 2 can still fail if we drop the completeness hypothesis on just X, or just Y.

Exercise 6. Suppose that is a surjective continuous linear transformation between Banach spaces.  By combining the open mapping theorem with the Hahn-Banach theorem, show that the transpose map is bounded from below, i.e. there exists such that for all .  Conclude that is an isomorphism between and . 

Let L be as in Theorem 3, so that the problem Lu=f is both qualitatively and quantitatively solvable.  A standard application of Zorn’s lemma (similar to that used to prove the Hahn-Banach theorem) shows that the problem Lu=f is also qualitatively linearly solvable, in the sense that there exists a linear transformation such that for all (i.e. S is a right-inverse of L).  In view of the open mapping theorem, it is then tempting to conjecture that L must also be quantitatively linearly solvable, in the sense that there exists a continuous linear transformation such that for all .  By Corollary 2, we see that this conjecture is true when the problem Lu=f is determined, i.e. there is exactly one solution u for each datum f.  Unfortunately, the conjecture can fail when Lu=f is underdetermined (more than one solution u for each f); we discuss this in the appendix to these notes.  On the other hand, the situation is much better for Hilbert spaces:

Exercise 7. Suppose that is a surjective continuous linear transformation between Hilbert spaces.  Show that there exists a continuous linear transformation such that .  Furthermore, we can ensure that the range of S is orthogonal to the kernel of L, and that this condition determines S uniquely.

Remark 7. In fact, Hilbert spaces are essentially the only type of Banach space for which we have this nice property, due to the Lindenstrauss-Tzafriri solution of the complemented subspaces problem.

Exercise 8. Let M and N be closed subspaces of a Banach space X.  Show that the following statements are equivalent:

1. (Qualitative complementation) Every x in X can be expressed in the form m+n for in exactly one way.
2. (Quantitative complementation)  Every x in X can be expressed in the form m+n for in exactly one way.  Furthermore there exists C > 0 such that all x.

When either of these two properties hold, we say that M (or N) is a complemented subspace, and that N is a complement of M (or vice versa). 

The property of being complemented is closely related to that of quantitative linear solvability:

Exercise 9. Let be a surjective bounded linear map between Banach spaces.  Show that there exists a bounded linear map such that for all if and only if the kernel is a complemented subspace of X. 

Exercise 10. Show that any finite-dimensional or closed finite co-dimensional subspace of a Banach space is complemented. 

Remark 8. The problem of determining whether a given closed subspace of a Banach space is complemented or not is, in general, quite difficult.  However, non-complemented subspaces do exist in abundance; some example are given in the apendix, and the Lindenstrauss-Tzafriri theorem referred to in in Remark 7 asserts that any Banach space not isomorphic to a Hilbert space contains at least one non-complemented subspace.  There is also a remarkable construction of Gowers and Maurey of a Banach space such that every subspace, other than those ruled out by Exercise 10, are uncomplemented. 

– The closed graph theorem –

Recall that a map between two metric spaces is continuous if and only if, whenever converges to x in X, converges to Tx in Y.  We can also define the weaker property of being closed: an map is closed if and only if whenever converges to x in X, and converges to a limit y in Y, then y is equal to Tx; equivalently, T is closed if its graph is a closed subset of .  This is weaker than continuity because it has the additional requirement that the sequence is already convergent. (Despite the name, closed operators are not directly related to open operators.)

Example 2. Let be the transformation .  This transformation is unbounded and hence discontinuous, but one easily verifies that it is closed. 

As Example 2 shows, being closed is often a weaker property than being continuous.  However, the remarkable closed graph theorem shows that as long as the domain and range of the operator are both Banach spaces, the two statements are equivalent:

Theorem 4. (Closed graph theorem)  Let be a linear transformation between two Banach spaces.  Then the following are equivalent:

1. T is continuous.
2. T is closed.
3. (Weak continuity) There exists some topology on Y, weaker than the norm topology (i.e. containing fewer open sets) but still Hausdorff, for which is continuous.

Proof. It is clear that 1 implies 3 (just take to equal the norm topology).  To see why 3 implies 2, observe that if in X and in norm, then in the weaker topology as well; but by weak continuity in .  Since Hausdorff topological spaces have unique limits, we have Tx=y and so T is closed.

Now we show that 2 implies 1.  If T is closed, then the graph is a closed linear subspace of the Banach space and is thus also a Banach space.  On the other hand, the projection map from to X is clearly a continuous linear bijection.  By Corollary 2, its inverse is also continuous, and so T is continuous as desired.

We can reformulate the closed graph theorem in the following fashion:

Corollary 3. Let X, Y be Banach spaces, and suppose we have some continuous inclusion of Y into a Hausdorff topological vector space Z.  Let be a continuous linear transformation.  Then the following are equivalent.

1. (Qualitative regularity) For all , .
2. (Quantitative regularity) For all , , and furthermore for some independent of x.
3. (Quantitative regularity on a dense subclass) For all x in a dense subset of X, , and furthermore for some independent of x.

Proof. Clearly 2. implies 3. or 1.  If we have 3., then T extends uniquely to a bounded linear map from X to Y, which must agree with the original continuous map from X to Z since limits in the Hausdorff space Z are unique, and so 3. implies 2.  Finally, if 1. holds, then we can view T as a map from X to Y, which by Theorem 4 is continuous, and the claim now follows from Lemma 1 from Notes 3.

In practice, one should think of Z as some sort of “low regularity” space with a weak topology, and Y as a “high regularity” subspace with a stronger topology.  Corollary 3 motivates the method of a priori estimates to establish the Y-regularity of some linear transform Tx of an arbitrary element x in a Banach space X, by first establishing the a priori estimate for a dense subclass of “nice” elements of X, and then using the above corollary (and some weak continuity of T in a low regularity space) to conclude.  The closed graph theorem provides the metamathematical explanation as to why this approach is at least as powerful as any other approach to proving regularity.

Example 3. Let , and let p’ be the dual exponent of p.  To prove that the Fourier transform of a function necessarily lies in , it suffices to prove the Hausdorff-Young inequality

(5)

for some constant and all f in some suitable dense subclass of (e.g. the space of smooth functions of compact support), together with the “soft” observation that the Fourier transform is continuous from to the space of tempered distributions, which is a Hausdorff space into which embeds continuously.  One can replace the Hausdorff-Young inequality here by countless other estimates in harmonic analysis to obtain similar qualitative regularity conclusions.

– Appendix: Nonlinear solvability (optional) –

In this appendix we give an example of a linear equations Lu=f which can only be quantitatively solved in a nonlinear fashion.  We will use a number of basic tools which we will only cover later in this course, and so this material is optional reading.

Let be the infinite discrete cube with the product topology; by Tychonoff’s theorem, this is a compact Hausdorff space.  The Borel -algebra is generated by the cylinder sets

. (6)

(From a probabilistic view point, one can think of X as the event space for flipping a countably infinite number of coins, and as the event that the coin lands as heads.)

Let be the space of finite Borel measures on X; this can be verified to be a Banach space.  There is a map defined by

. (7)

This is a continuous linear transformation.  The equation is quantitatively solvable for every .  Indeed, if f is an indicator function , then , where is the sequence that equals 1 on A and 0 outside of A, and is the Dirac mass at A.  The general case then follows by expressing a bounded sequence as an integral of indicator functions (e.g. if f takes values in [0,1], we can write ).  Note however that this is a nonlinear operation, since the indicator depends nonlinearly on f.

We now claim that the equation is not quantitatively linearly solvable, i.e. there is no bounded linear map such that LSf = f for all .  This fact was first observed by Banach and Mazur; we shall give two proofs, one of a “soft analysis” flavour and one of a “hard analysis” flavour.

We begin with the “soft analysis” proof, starting with a measure-theoretic result which is of independent interest.

Theorem 5. (Nikodym convergence theorem) Let be a measurable space, and let be a sequence of signed finite measures which is weakly convergent in the sense that converges to some limit for each .

1. The are uniformly countably additive, which means that for any sequence of disjoint measurable sets, the series converges uniformly in n.
2. is a signed finite measure.

Proof. It suffices to prove the first part, since this easily implies that is also countably additive, and is thence a signed finite measure.  Suppose for contradiction that the claim failed, then one could find disjoint and such that one has for all M.  We now construct disjoint sets , each consisting of the union of a finite collection of the , and an increasing sequence of positive integers, by the following recursive procedure:

1. Initialise .
2. Suppose recursively that and has already been constructed for some .
3. Choose so large that for all , differs from by at most .
4. Choose so large that is larger than j for any , and such that for all .
5. Choose so that .
6. Pick to be a finite union of the with such that .
7. Increment k to k+1 and then return to Step 2.

It is then a routine matter to show that if , then for all j, contradicting the hypothesis that is weakly convergent to .

Exercise 11. (Schur’s property for )  Show that if a sequence in is convergent in the weak topology, then it is convergent in the strong topology. 

We return now to the map .  Consider the sequence defined by , i.e. each is the sequence consisting of n 1′s followed by an infinite number of 0′s.   As the dual of is isomorphic to , we see from the dominated convergence theorem that is a weakly Cauchy sequence in , in the sense that is Cauchy for any .  Applying S, we conclude that is weakly Cauchy in .  In particular, using the bounded linear functionals on M(X), we see that converges to some limit for all measurable sets E.  Applying the Nikodym convergence theorem we see that is also a signed finite measure.  We then see that converges in the weak topology to .  (One way to see this is to define , then is finite and are all absolutely continuous with respect to ; now use the Radon-Nikodym theorem (see Notes 1) and the fact that .)  On the other hand, as and L and S are both bounded, S is a Banach space isomorphism between and .  Thus is complete, hence closed, hence weakly closed (by Hahn-Banach), and so for some .  By Hahn-Banach again, this implies that converges weakly to .  But this is easily seen to be impossible, since the constant sequence does not lie in , and the claim follows.

Now we give the “hard analysis” proof.  Let be the standard basis for , let N be a large number, and consider the random sums

(8)

where are iid random signs.  Since the norm of is 1, we have

(9)

for some constant C independent of N.  On the other hand, we can write for some finite measure and some using  Radon-Nikodym as in the previous proof, and then

(10)

Taking expectations and applying Khintchine’s inequality we conclude

(11)

for some constant C’ independent of N.  By Cauchy-Schwarz this implies that

(12)

But as for some constant c > 0 independent of N, we obtain a contradiction for N large enough, and the claim follows.

Remark 9. The phenomenon of nonlinear quantitative solvability actually comes up in many applications of interest.  For instance, consider the Fefferman-Stein decomposition theorem, which asserts that any of bounded mean oscillation can be decomposed as for some , where H is the Hilbert transform.  This theorem was first proven by using the duality of the Hardy space and BMO (and by using Exercise 13 from Notes 6), and by using the fact that a function f is in if and only if f and Hf both lie in .  From the open mapping theorem we know that we can pick g, h so that the norms of g, h are bounded by a multiple of the BMO norm of f.  But it turns out not to be possible to pick g and h in a bounded linear manner in terms of f, although this is a little tricky to prove.  (Uchiyama famously gave an explicit construction of g, h in terms of f, but the construction was highly nonlinear; see my blog post on the topic.)

An example in a similar spirit was given more recently by Bourgain and Brezis, who considered the problem of solving the equation on the d-dimensional torus for some function on the torus with mean zero, and with some unknown vector field , where the derivatives are interpreted in the weak sense.  They showed that if and , then there existed a solution u to this problem with , despite the failure of Sobolev embedding at this endpoint.  Again, the open mapping theorem allows one to choose u with norm bounded by a multiple of the norm of f, but Bourgain and Brezis also show that one cannot select u in a bounded linear fashion depending on f. 

Question. All of the above constructions of non-complemented closed subspaces, or of linear problems that can only be quantitatively solved nonlinearly, were quite involved.  Is there a “soft” or “elementary” way to see that closed subspaces of Banach spaces exist which are not complemented, or (equivalently) that surjective continuous linear maps between Banach spaces do not always enjoy a continuous linear right-inverse?  I do not have a good answer to this question.

[Update, Feb 4: definition of "residual" corrected.]

– Metric spaces –

In many spaces, one wants a notion of when two points in the space are “near” or “far”.  A particularly quantitative and intuitive way to formalise this notion is via the concept of a metric space.

Definition 1. (Metric spaces)  A metric space is a set X, together with a distance function which obeys the following properties:

1. (Non-degeneracy) For any , we have , with equality if and only if x=y.
2. (Symmetry) For any , we have .
3. (Triangle inequality) For any , we have .

Example 1. Every normed vector space is a metric space, with distance function .

Example 2. Any subset Y of a metric space is also a metric space , where is the restriction of d to .  We call the metric space a subspace of the metric space .

Example 3. Given two metric spaces and , we can define the product space to be the Cartesian product with the product metric

. (1)

(One can also pick slightly different metrics here, such as , but this metric only differs from (1) by a factor of two, and so they are equivalent (see Example 5 below). 

Example 4. Any set X can be turned into a metric space by using the discrete metric , defined by setting when and otherwise.

Given a metric space, one can then define various useful topological structures.  There are two ways to do so.  One is via the machinery of convergent sequences:

Definition 2. (Topology of a metric space)  Let be a metric space.

1. A sequence of points in X is said to converge to a limit if one has as .  In this case, we say that in the metric d as , and that in the metric space X.  (It is easy to see that any sequence of points in a metric space has at most one limit.)
2. A point x is an adherent point of a set if it is the limit of some sequence in E.  (This is slightly different from being a limit point of E, which is equivalent to being an adherent point of ; every adherent point is either a limit point or an isolated point of E.)  The set of all adherent points of E is called the closure of X.  A set E is closed if it contains all its adherent points, i.e. if .  A set E is dense if every point in X is adherent to E, or equivalently if .
3. Given any x in X and , define the open ball centred at x with radius r to be the set of all y in X such that .  Given a set E, we say that x is an interior point of E if there is some open ball centred at x which is contained in E.  The set of all interior points is called the interior of E.  A set is open if every point is an interior point, i.e. if .

There is however an alternate approach to defining these concepts, which takes the concept of an open set as a primitive, rather than the distance function, and defines other terms in terms of open sets.  For instance:

Exercise 1. Let be a metric space.

1. Show that a sequence of points in X converges to a limit if and only if every open neighbourhood of x (i.e. an open set containing x) contains for all sufficiently large n.
2. Show that a point x is an adherent point of a set E if and only if every open neighbourhood of x intersects E.
3. Show that a set E is closed if and only if its complement is open.
4. Show that the closure of a set E is the intersection of all the closed sets containing E.
5. Show that a set E is dense if and only if every non-empty open set intersects E.
6. Show that the interior of a set E is the union of all the open sets contained in E, and that x is an interior point of E if and only if some neighbourhood of x is contained in E.

In the next section we will adopt this “open sets first” perspective when defining topological spaces.

On the other hand, there are some other properties of subsets of a metric space which require the metric structure more fully, and cannot be defined purely in terms of open sets (see Example 14) below (although some of these concepts can still be defined using a structure intermediate to metric spaces and topological spaces, namely a uniform space).  For instance:

Definition 3. Let (X,d) be a metric space.

1. A sequence of points in X is a Cauchy sequence if as (i.e. for every there exists such that for all ).
2. A space X is complete if every Cauchy sequence is convergent.
3. A set E in X is bounded if it is contained inside a ball.
4. A set E is totally bounded in X if for every , E can be covered by finitely many balls of radius .

Exercise 2. Show that any metric space can be identified with a dense subspace of a complete metric space , known as a metric completion or Cauchy completion of X.  (For instance, is a metric completion of .)  (Hint: one can define a real number to be an equivalence class of Cauchy sequences of rationals.  Once the reals are defined, essentially the same construction works in arbitrary metric spaces.) Furthermore, if is another metric completion of , show that there exists an isometry between and which is the identity on X.  Thus, up to isometry, there is a unique metric completion to any metric space. 

Exercise 3. Show that a metric space X is complete if and only if it is closed in every superspace Y of X (i.e. in every metric space Y for which X is a subspace).  Thus one can think of completeness as being the property of being “absolutely closed”.

Exercise 4. Show that every totally bounded set is also bounded.  Conversely, in a Euclidean space with the usual metric, show that every bounded set is totally bounded.  But give an example of a set in a metric space which is bounded but not totally bounded.  (Hint: use Example 4.)

Now we come to an important concept.

Theorem 1. (Heine-Borel theorem for metric spaces)  Let be a metric space.  Then the following are equivalent:

1. (Sequential compactness) Every sequence in X has a convergent subsequence.
2. (Compactness) Every open cover of X (i.e. a collection of open sets whose union contains X) has a finite subcover.
3. (Finite intersection property)  If is a collection of closed subsets of X such that any finite subcollection of sets has non-empty intersection, then the entire collection has non-empty intersection.
4. X is complete and totally bounded.

Proof. (2 1) If there was an infinite sequence with no convergent subsequence, then given any point x in X there must exist an open ball centred at x which contains for only finitely many n (since otherwise one could easily construct a subsequence of converging to x).  By property 2, one can cover X with a finite number of such balls.  But then the sequence would be finite, a contradiction.

(1 4)  If X was not complete, then there would exist a Cauchy sequence which is not convergent; one easily shows that this sequence cannot have any convergent subsequences either, contradicting 1.  If X was not totally bounded, then there exists such that X cannot be covered by any finite collection of balls of radius ; a standard greedy algorithm argument then gives a sequence such that for all distinct n, m.  This sequence clearly has no convergent subsequence, again a contradiction.

(2 3)  This follows from de Morgan’s laws and Exercise 1.3.

(4 3)  Let be as in 3.  Call a set E in X rich if it intersects all of the .  Observe that if one could cover X by a finite number of non-rich sets, then (as each non-rich set is disjoint from at least one of the ), there would be a finite number of whose intersection is empty, a contradiction.  Thus, whenever we cover X by finitely many sets, at least one of them must be rich.

As X is totally bounded, for each we can find a finite set such that the balls cover X.  By the previous discussion, we can then find such that is rich.

Call a ball asymptotically rich if it contains infinitely many of the .  As these balls cover X, we see that for each n, is asymptotically rich for at least one i.  Furthermore, since each ball of radius can be covered by balls of radius , we see that if  is asymptotically rich, then it must intersect an asymptotically rich ball .  Iterating this, we can find a sequence of asymptotically rich balls, each one of which intersects the next one.  This implies that is a Cauchy sequence and hence (as X is assumed complete) converges to a limit x.  Observe that there exist arbitrarily small rich balls that are arbitrarily close to x, and thus x is adherent to every ; since the are closed, we see that x lies in every , and we are done. 

Remark 1. The hard implication of the Heine-Borel theorem is noticeably more complicated than any of the others.  This turns out to be unavoidable; the Heine-Borel theorem turns out to be logically equivalent to König’s lemma in the sense of reverse mathematics, and thus cannot be proven in sufficiently weak systems of logical reasoning. 

Any space that obeys one of the four equivalent properties in Lemma 1 is called a compact space; a subset E of a metric space X is said to be compact if it is a compact space when viewed as a subspace of X. There are some variants of the notion of compactness which are also of importance for us:

1. A space is -compact if it can be expressed as the countable union of compact sets.  (For instance, the real line with the usual metric is -compact.)
2. A space is locally compact if every point is contained in the interior of a compact set.  (For instance, is locally compact.)
3. A subset of a space is precompact or relatively compact if it is contained inside a compact set (or equivalently, if its closure is compact).

Another fundamental notion in the subject is that of a continuous map.

Exercise 5. Let be a map from one metric space to another .  Then the following are equivalent:

1. (Metric continuity) For every and there exists such that whenever .
2. (Sequential continuity) For every sequence that converges to a limit , converges to f(x).
3. (Topological continuity) The inverse image of every open set V in Y, is an open set in X.
4. The inverse image of every closed set F in Y, is a closed set in X. 

A function f obeying any one of the properties in Exercise 5 is known as a continuous map.

Exercise 6. Let be metric spaces, and let and be continuous maps.  Show that the combined map defined by is continuous if and only if f and g are continuous.  Show also that the projection maps , defined by , are continuous. 

Exercise 7. Show that the image of a compact set under a continuous map is again compact.

– Topological spaces –

Metric spaces capture many of the notions of convergence and continuity that one commonly uses in real analysis, but there are several such notions (e.g. pointwise convergence, semi-continuity, or weak convergence) in the subject that turn out to not be modeled by metric spaces.  A very useful framework to handle these more general modes of convergence and continuity is that of a topological space, which one can think of as an abstract generalisation of a metric space in which the metric and balls are forgotten, and the open sets become the central object.  [There are even more abstract notions, such as pointless topological spaces, in which the collection of open sets has become an abstract lattice, in the spirit of Notes 4, but we will not need such notions in this course.]

Definition 4. (Topological space)  A topological space is a set X, together with a collection of subsets of X, known as open sets, which obey the following axioms:

1. and X are open.
2. The intersection of any finite number of open sets is open.
3. The union of any arbitrary number of open sets is open.

The collection is called a topology on X.

Given two topologies on a space X, we say that is a coarser (or weaker) topology than (or equivalently, that is a finer (or stronger) topology than ), if (informally, has more open sets than ).

Example 5. Every metric space generates a topology , namely the space of sets which are open with respect to the metric d.  Observe that if two metrics d, d’ on X are equivalent in the sense that

(2)

for all x, y in X and some constants , then they generate an identical topology.

Example 6. The finest (or strongest) topology on any set X is the discrete topology , in which every set is open; this is the topology generated by the discrete metric (Example 4).  The coarsest (or weakest) topology is the trivial topology , in which only the empty set and the full set are open. 

Example 7. Given any collection of sets of X, we can define the topology generated by to be the intersection of all the topologies that contain ; this is easily seen to be the coarsest topology that makes all the sets in open.  For instance, the topology generated by a metric space is the same as the topology generated by its open balls.

Example 8. If is a topological space, and Y is a subset of X, then we can define the relative topology to be the collection of all open sets in X, restricted to Y, this makes a topological space, known as a subspace of . 

Any notion in metric space theory which can be defined purely in terms of open sets, can now be defined for topological spaces.  Thus for instance:

Definition 5. Let be a topological space.

1. A sequence of points in X converges to a limit if and only if every open neighbourhood of x (i.e. an open set containing x) contains for all sufficiently large n.  In this case we write in the topological space , and (if x is unique) we write .
2. A point is a sequentially adherent point of a set E if it is the limit of some sequence in E.
3. A point x is an adherent point of a set E if and only if every open neighbourhood of x intersects E.  The set of all adherent points of E is called the closure of E and is denoted .
4. A set E is closed if and only if its complement is open, or equivalently if it contains all its adherent points.
5. A set E is dense if and only if every non-empty open set intersects E, or equivalently if its closure is X.
6. The interior of a set E is the union of all the open sets contained in E, and x is called an interior point of E if and only if some neighbourhood of x is contained in E.
7. A space X is sequentially compact if every sequence has a convergent subsequence.
8. A space X is compact if every open cover has a finite subcover.
9. The concepts of being -compact, locally compact, and precompact can be defined as before.  (One could also define sequential -compactness, etc., but these notions are rarely used.)
10. A map between topological spaces is sequentially continuous if whenever converges to a limit x in X, converges to a limit f(x) in Y.
11. A map between topological spaces is continuous if the inverse image of every open set is open.

Remark 2. The stronger a topology becomes, the more open and closed sets it will have, but fewer sequences will converge, there are fewer (sequentially) adherent points and (sequentially) compact sets, closures become smaller, and interiors become larger.  There will be more (sequentially) continuous functions on this space, but fewer (sequentially) continuous functions into the space.   Note also that the identity map from a space X with one topology to the same space X with a different topology is continuous precisely when is stronger than . 

Example 9. In a metric space, these topological notions coincide with their metric counterparts, and sequential compactness and compactness are equivalent, as are sequential continuity and continuity. 

Exercise 7′. (Urysohn’s subsequence principle) Let be a sequence in a topological space X, and let x be another point in X.  Show that the following are equivalent:

1. converges to x.
2. Every subsequence of converges to x.
3. Every subsequence of has a further subsequence that converges to x.

Exercise 8. Show that every sequentially adherent point is an adherent point, every continuous function is sequentially continuous.

Remark 3. The converses to Exercise 8 are unfortunately not always true in general topological spaces.  For instance, if we endow an uncountable set X with the cocountable topology (so that a set is open if it is either empty, or its complement is at most countable) then we see that the only convergent sequences are those which are eventually constant.  Thus, every subset of X contains its sequentially adherent points, and every function from X to another topological space is sequentially continuous, even though not every set in X is closed and not every function on X is continuous.  An example of a set which is sequentially compact but not compact is the first uncountable ordinal with the order topology (Exercise 9).  It is more tricky to give an example of a compact space which is not sequentially compact; this will have to wait for future notes, when we establish Tychonoff’s theorem.  However one can “fix” this discrepancy between the sequential and non-sequential concepts by replacing sequences with the more general notion of nets, see the appendix below.

Remark 4. Metric space concepts such as boundedness, completeness, Cauchy sequences, and uniform continuity do not have counterparts for general topological spaces, because they cannot be defined purely in terms of open sets.  (They can however be extended to some other types of spaces, such as uniform spaces or coarse spaces.) 

Now we give some important topologies that capture certain modes of convergence or continuity that are difficult or impossible to capture using metric spaces alone.

Example 10. (Zariski topology)  This topology is important in algebraic geometry, though it will not be used in this course.  If F is an algebraically closed field, we define the Zariski topology on the vector space to be the topology generated by the complements of proper algebraic varieties in ; thus a set is Zariski open if it is either empty, or is the complement of a finite union of proper algebraic varieties.  A set in is then Zariski dense if it is not contained in any proper subvariety, and the Zariski closure of a set is the smallest algebraic variety that contains that set. 

Example 11. (Order topology)  Any totally ordered set generates the order topology, defined as the topology generated by the sets and for all .  In particular, the extended real line can be given the order topology, and the notion of convergence of sequences in this topology to either finite or infinite limits is identical to the notion one is accustomed to in undergraduate real analysis.  (On the real line, of course, the order topology corresponds to the usual topology.)  Also observe that a function from the extended natural numbers (with the order topology) into a topological space X is continuous if and only if as , so one can interpret convergence of sequences as a special case of continuity.

Exercise 9. Let be the first uncountable ordinal, endowed with the order topology.  Show that is sequentially compact (Hint: every sequence has a lim sup), but not compact (Hint: every point has a countable neighbourhood).

Example 12. (Half-open topology)  The right half-open topology on the real line is the topology generated by the right half-open intervals for ; this is a bit finer than the usual topology on .  Observe that a sequence converges to a limit x in the right half-open topology if and only if it converges in the ordinary topology , and also if for all sufficiently large x.  Observe that a map is right-continuous iff it is a continuous map from to .  One can of course model left-continuity via a suitable left half-open topology in a similar fashion.

Example 13. (Upper topology)  The upper topology on the real line is defined as the topology generated by the sets for all .  Observe that (somewhat confusingly), a function is lower semi-continuous iff it is continuous from to . One can of course model upper semi-continuity via a suitable lower topology in a similar fashion. 

Example 14. (Product topology)  Let be the space of all functions from a set X to a topological space Y.  We define the product topology on to be the topology generated by the sets for all and all open .  Observe that a sequence of functions converges pointwise to a limit iff it converges in the product topology.  We will study the product topology in more depth in future notes.

Example 15. (Product topology, again) If and are two topological spaces, we can define the product space to be the Cartesian product with the topology generated by the product sets , where U and V are open in X and Y respectively.  Observe that two functions , from a topological space Z are continuous if and only if their direct sum is continuous in the product topology, and also that the projection maps and are continuous (cf. Exercise 6). 

We mention that not every topological space can be generated from a metric (such topological spaces are called metrisable).  One important obstruction to this arises from the Hausdorff property:

Definition 6. A topological space X is said to be a Hausdorff space if for any two distinct points x, y in X, there exist disjoint neighbourhoods of x and y respectively.

Example 16. Every metric space is Hausdorff (one can use the open balls and as the separating neighbourhoods.  On the other hand, the trivial topology (Example 7)  on two or more points is not Hausdorff, and neither is the cocountable topology (Remark 3) on an uncountable set, or the upper topology (Example 13) on the real line.   Thus, these topologies do not arise from a metric.

Exercise 10. Show that the half-open topology (Example 12) is Hausdorff, but does not arise from a metric.  [Hint: assume for contradiction that the half-open topology did arise from a metric; then show that for every real number x there exists a rational number q and a positive integer n such that the ball of radius 1/n centred at q has infimum x.]  Thus there are more obstructions to metrisability than just the Hausdorff property; a more complete answer is provided by Urysohn’s metrisation theorem, which we will cover in later notes.

Exercise 11. Show that in a Hausdorff space, any sequence can have at most one limit.  (For a more precise statement, see Exercise 15 below.)

A homeomorphism (or topological isomorphism) between two topological spaces is a continuous invertible map whose inverse is also continuous.  Such a map identifies the topology on X with the topology on Y, and so any topological concept of X will be preserved by f to the corresponding topological concept of Y.  For instance, X is compact if and only if Y is compact, X is Hausdorff if and only if Y is Hausdorff, x is adherent to E if and only if f(x) is adherent to f(E), and so forth.  When there is a homeomorphism between two topological spaces, we say that X and Y are homeomorphic (or topologically isomorphic).

Example 14. The tangent function is a homeomorphism between and (with the usual topologies), and thus preserves all topological structures on these two spaces.  Note however that the former space is bounded as a metric space while the latter is not, and the latter is complete while the former is not.  Thus metric properties such as boundedness or completeness are not purely topological properties, since they are not preserved by homeomorphisms.

– Nets (optional) –

A sequence in a space X can be viewed as a function from the natural numbers to X.  We can generalise this concept as follows.

Definition 7. A net in a space X is a tuple , where is a directed set (i.e. a partially ordered set such that any two elements have at least one upper bound), and for each .  We say that a statement holds for sufficiently large in a directed set A if there exists such that holds for all .  [Note in particular that if and separately hold for sufficiently large , then their conjunction also holds for sufficiently large .]

A net in a topological space X is said to converge to a limit if for every neighbourhood V of x, we have for all sufficiently large .

A subnet of a net is a tuple of the form , where is another directed set, and is a monotone map (thus whenever ) which is also has cofinal image, which means that for any there exists with (in particular, if is true for sufficiently large , then is true for sufficiently large ).

Remark 5. Every sequence is a net, but one can create nets that do not arise from sequences (in particular, one can take A to be uncountable). Note a subtlety in the definition of a subnet – we do not require to be injective, so B can in fact be larger than A!  Thus subnets differ a little bit from subsequences in that they “allow repetitions”.

Remark 6. Given a directed set A, one can endow with the topology generated by the singleton sets with , together with the sets for , with the convention that for all .  The property of being directed is precisely saying that these sets form a base.  A net converges to a limit if and only if the function is continuous on (cf. Example 11).  Also, if is a subnet of , then is a continuous map from to , if we adopt the convention that .  In particular, a subnet of a convergent net remains convergent to the same limit.

The point of working with nets instead of sequences is that one no longer needs to worry about the distinction between sequential and non-sequential concepts in topology, as the following exercises show:

Exercise 12. Let X be a topological space, let E be a subset of X, and let x be an element of X.  Show that x is an adherent point of E if and only if there exists a net in E that converges to x.  (Hint: take A to be the directed set of neighbourhoods of x, ordered by reverse set inclusion.)

Exercise 13. Let be a map between two topological spaces.  Show that f is continuous if and only if for every net in X that converges to a limit x, the net converges in Y to f(x).

Exercise 14. Let X be a topological space.  Show that X is compact if and only if every net has a convergent subnet.  (Hint: equate both properties of X with the finite intersection property, and review the proof of Theorem 1.    Similarly, show that a subset E of X is relatively compact if and only if every net in E has a subnet that converges in X.  (Note that as not every compact space is sequentially compact, this exercise shows that we cannot enforce injectivity of in the definition of a subnet.)

Exercise 15. Show that a space is Hausdorff if and only if every net has at most one limit.

Exercise 16. In the product space in Example 14, show that a net converges in to if and only if for every , the net converges in Y to .

[Update, Jan 31: Definition of subnet corrected; Exercise 8 corrected; Exercise 9 added, subsequent exercises renumbered; hint for Exercise 2 altered; some remarks added.]

In this course we will often need to iterate some sort of operation “infinitely many times” (e.g. to create a infinite basis by choosing one basis element at a time).  In order to do this rigorously, we will rely on Zorn’s lemma:

Zorn’s Lemma. Let be a non-empty partially ordered set, with the property that every chain (i.e. a totally ordered set) in X has an upper bound.  Then X contains a maximal element (i.e. an element with no larger element).

Indeed, we have used this lemma several times already in previous notes.  Given the other standard axioms of set theory, this lemma is logically equivalent to

Axiom of choice. Let X be a set, and let be a collection of non-empty subsets of X.  Then there exists a choice function , i.e. a function such that for all .

One implication is easy:

Proof of axiom of choice using Zorn’s lemma. Define a partial choice function to be a pair , where is a subset of and is a choice function for .  We can partially order the collection of partial choice functions by writing if and f” extends f’.  The collection of partial choice functions is non-empty (since it contains the pair consisting of the empty set and the empty function), and it is easy to see that any chain of partial choice functions has an upper bound (formed by gluing all the partial choices together).  Hence, by Zorn’s lemma, there is a maximal partial choice function .  But the domain of this function must be all of , since otherwise one could enlarge by a single set A and extend to A by choosing a single element of A.  (One does not need the axiom of choice to make a single choice, or finitely many choices; it is only when making infinitely many choices that the axiom becomes necessary.)  The claim follows.

In the rest of these notes I would like to supply the reverse implication, using the machinery of well-ordered sets.  Instead of giving the shortest or slickest proof of Zorn’s lemma here, I would like to take the opportunity to place the lemma in the context of several related topics, such as ordinals and transfinite induction, noting that much of this material is in fact independent of the axiom of choice.  The material here is standard, but for the purposes of this course one may simply take Zorn’s lemma as a “black box” and not worry about the proof, so this material is optional.

– Well-ordered sets –

To prove Zorn’s lemma, we first need to strengthen the notion of a totally ordered set.

Definition 1. A well-ordered set is a totally ordered set such that every non-empty subset A of X has a minimal element .  Two well-ordered sets X, Y are isomorphic if there is an order isomorphism between them, i.e. a bijection which is monotone ( whenever ).

Example 1. The natural numbers are well-ordered (this is the well-ordering principle), as is any finite totally ordered set (including the empty set), but the integers, rationals, or reals are not well-ordered.

Example 2. Any subset of a well-ordered set is again well-ordered.  In particular, if a, b are two elements of a well-ordered set, then intervals such as , , etc. are also well-ordered.

Example 3. If X is a well-ordered set, then the ordered set , defined by adjoining a new element to and declaring it to be larger than all the elements of X, is also well-ordered.  More generally, if X and Y are well-ordered sets, then the ordered set , defined as the disjoint union of X and Y, with any element of Y declared to be larger than any element of X, is also well-ordered.  Observe that the operation is associative (up to isomorphism), but not commutative in general: for instance, is not isomorphic to .

Example 4. If X, Y are well-ordered sets, then the ordered set , defined as the Cartesian product with the lexicographical ordering (thus if , or if and ), is again a well-ordered set.  Again, this operation is associative (up to isomorphism) but not commutative.  Note that we have one-sided distributivity: is isomorphic to , but is not isomorphic to in general.

Remark 1. The axiom of choice is trivially true in the case when X is well-ordered, since one can take to be the choice function.  Thus, the axiom of choice follows from the well-ordering theorem (every set has at least one well-ordering).  Conversely, we will be able to deduce the well-ordering theorem from Zorn’s lemma (and hence from the axiom of choice): see Exercise 11 below.

One of the reasons that well-ordered sets are useful is that one can perform induction on them.  This is easiest to describe for the principle of strong induction:

Exercise 1. (Strong induction on well-ordered sets)  Let X be a well-ordered set, and let be a property of elements of X.  Suppose that whenever is such that is true for all , then is true.  Then is true for every . This is called the principle of strong induction.  Conversely, show that a totally ordered set X enjoys the principle of strong induction if and only if it is well-ordered.  (For partially ordered sets, the corresponding notion is that of being well-founded.)

To describe the analogue of the ordinary principle of induction for well-ordered sets, we need some more notation.  Given a subset A of a non-empty well-ordered set X, we define the supremum of A to be the least upper bound

(1)

of A (thus for instance the supremum of the empty set is ).  If , we define the successor by the formula

. (2)

We have the following Peano-type axioms:

Exercise 2. If x is an element of a non-empty well-ordered set X, show that exactly one of the following statements hold:

1. (Limit case) .
2. (Successor case) for some .

In particular, is not the successor of any element in X.

Exercise 3. Show that if x, y are elements of a well-ordered set such that , then x=y.

Exercise 4. (Transfinite induction for well-ordered sets)   Let X be a non-empty well-ordered set, and let be a property of elements of X. Suppose that

1. (Base case) is true.
2. (Successor case) If and is true, then is true.
3. (Limit case) If and is true for all , then is true.  [Note that this subsumes the base case.]

Then is true for all . 

Remark 2. The usual Peano axioms for succession are the special case of Exercises 2-4 in which the limit case of Exercise 2 only occurs for (which is denoted 0), and the successor function never attains .  With these additional axioms, X is necessarily isomorphic to . 

Now we introduce two more key concepts.

Definition 2. An initial segment of a well-ordered set X is a subset Y of X such that for all (i.e. whenever y lies in Y, all elements of X that are less than y also lie in Y).

A morphism from one well-ordered set X to another Y is a map which is strictly monotone (thus whenever ) and such that is an initial segment of Y.

Example 1. The only morphism from to is the inclusion map.  There is no morphism from to . 

Remark 3. With this notion of a morphism, the class of well-ordered sets becomes a category.

We can identify the initial segments of X with elements of :

Exercise 5. Let X be a non-empty well-ordered set.  Show that every initial segment I of X is of the form for exactly one .

Exercise 6. Show that an arbitrary union or arbitrary intersection of initial segments is again an initial segment.

Exercise 7. Let be a morphism.  Show that maps initial segments of X to initial segments of Y.  If is such that x’ is the successor of x, show that is the successor of . 

As Example 1 suggests, there are very few morphisms between well-ordered sets.  Indeed, we have

Proposition 1. (Uniqueness of morphisms) Given two well-ordered sets X and Y, there is at most one morphism from X and Y.

Proof. Suppose we have two morphisms , .  By using transfinite induction (Exercise 4) and Exercise 7, one can show that agree on for every ; setting gives the claim. 

Exercise 8. (Schroder-Bernstein theorem for well-ordered sets) Show that two well-ordered sets X, Y are isomorphic if and only if there is a morphism from X to Y, and a morphism from Y to X. 

We can complement the uniqueness in Proposition 1 with existence:

Proposition 2. (Existence of morphisms) Given two well-ordered sets X and Y, there is either a morphism from X to Y or a morphism from Y to X.

Proof. Call an element good if there is a morphism from to Y, thus is good.  If is good, then we are done.  From uniqueness we see that if every element in a set A is good, then the supremum is also good.  Applying transfinite induction (Exercise 4), we thus see that we are done unless there exists a good such that is not good.  By Exercise 5, for some .  If then we could extend the morphism to by mapping a to b, contradicting the fact that is not good; thus and so is surjective.  It is then easy to check that exists and is a morphism from Y to X, and the claim follows.

Remark 4. Formally, Proposition 1, Exercise 8, and Proposition 2 tell us that the collection of all well-ordered sets, modulo isomorphism, is totally ordered by declaring one well-ordered set X to be at least as large as another Y when there is a morphism from Y to X.  However, this is not quite the case, because the collection of well-ordered sets is only a class rather than a set.  Indeed, as we shall soon see, this is not a technicality, but is in fact a fundamental fact about well-ordered sets that lies at the heart of Zorn’s lemma.  (From Russell’s paradox we know that the notions of class and set are necessarily distinct.) 

– Ordinals –

As we learn very early on in our mathematics education, a finite set of a certain cardinality (e.g. a set ) can be put in one-to-one correspondence with a “standard” set of the same cardinality (e.g. the set ); two finite sets have the same cardinality if and only if they correspond to the same “standard” set ).  (The same fact is true for infinite sets; see Exercise 12 below.) Similarly, we would like to place every well-ordered set in a “standard” form.  This motivates

Definition 3. A representation of the well-ordered sets is an assignment of a well-ordered set to every well-ordered set X such that

1. is isomorphic to X for every well-ordered set X.  (In particular, if and are equal, then X and Y are isomorphic.)
2. If there exists a morphism from X to Y, then is a subset of (and the order structure on is induced from that on .  (In particular, if X and Y are isomorphic, then and are equal.)

Remark 5. In the language of category theory, a representation is a covariant functor from the category of well-ordered sets to itself which turns all morphisms into inclusions, and which is naturally isomorphic to the identity functor.

Remark 6. Because the collection of all well-ordered sets is a class rather than a set, is not actually a function (it is sometimes referred to as a class function).  

It turns out that several representations of the well-ordered sets exist.  The most commonly used one is that of the ordinals, defined by von Neumann as follows.

Definition 4. An ordinal is a well-ordered set with the property that for all .  (In particular, each element of is also a subset of , and the strict order relation on is identical to the set membership relation .)

Example 5. For each natural number , define the ordinal number recursively by setting and for all , thus for instance