complete-lattices &laquo; WordPress.com Tag Feed http://en.wordpress.com/tag/complete-lattices/ Feed of posts on WordPress.com tagged "complete-lattices" Thu, 23 May 2013 06:27:50 +0000 http://en.wordpress.com/tags/ en <![CDATA[Strong vs weak partitioning - counterexample]]> http://portonmath.wordpress.com/2010/02/19/strong-weak-paritition-solved/ Fri, 19 Feb 2010 20:01:37 +0000 porton http://portonmath.wordpress.com/2010/02/19/strong-weak-paritition-solved/ My problem whether weak partitioning and strong partitioning of a complete lattice are the same was solved by counter-example by François G. Dorais.

Remains open whether strong and weak partitioning of a filter object is the same.

]]> <![CDATA[Chain-meet-closed sets on complete lattices]]> http://portonmath.wordpress.com/2009/12/12/chain-meet-closed/ Sat, 12 Dec 2009 14:13:20 +0000 porton http://portonmath.wordpress.com/2009/12/12/chain-meet-closed/ Let \mathfrak{A} is a complete lattice. I will call a filter base a nonempty subset T of \mathfrak{A} such that \forall a,b\in T\exists c\in T: (c\le a\wedge c\le b). I will call a chain (on \mathfrak{A}) a linearly ordered subset of \mathfrak{A}.

Now as a part my research of filters I attempt to solve this problem (the problem seems not very difficult and I hope to prove it today or tomorrow, however who knows how difficult it may be):

Definition A subset S of a complete lattice \mathfrak{A} is chain-meet-closed iff for every non-empty chain T\in\mathscr{P}S we have \bigcap T\in S.

Conjecture A subset S of a complete lattice \mathfrak{A} is chain-meet-closed iff for every filter base T\in\mathscr{P}S we have \bigcap T\in S.

]]> <![CDATA[Complete lattice generated by a partitioning - finite meets]]> http://portonmath.wordpress.com/2009/10/20/generated-lattice-finite-meets/ Tue, 20 Oct 2009 15:14:04 +0000 porton http://portonmath.wordpress.com/2009/10/20/generated-lattice-finite-meets/ I conjectured certain formula for the complete lattice generated by a strong partitioning of an element of complete lattice. Now I have found a beautiful proof of a weaker statement than this conjecture. (Well, my proof works only in the case of distributive lattices, but the case of non-distributive lattices is outside of my research area.)

Let’s denote R = \left\{ \bigcup{}^{\mathfrak{A}}X | X\in\mathscr{P}S \right\} where S is a strong partitioning an element of the complete lattice \mathfrak{A}. Our conjecture is trivially equivalent to the statement that R is closed under arbitrary meets and joins.

That R is closed regarding any joins is obvious. To finish proving the conjecture we need to show that R is closed under arbitrary meets. In this post I prove weaker result that R is closed under finite meets.

I hope this finite case may serve as a model for the general infinite case. However it seems that generalizing it to infinite case is non-trivial.

Theorem Let \mathfrak{A} is a distributive complete lattice and S is a strong partitioning of some element of this lattice. Then R is closed under finite meets.

Proof Let X,Y\in\mathscr{P}S.

Then \bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} ((X \cap Y) \cup (X \setminus Y)) \cap^{\mathfrak{A}}\bigcup^{\mathfrak{A}} Y = ( \bigcup^{\mathfrak{A}} (X \cap Y)\cup^{\mathfrak{A}} \bigcup^{\mathfrak{A}} (X \setminus Y))\cap^{\mathfrak{A}} \bigcup Y = ( \bigcup^{\mathfrak{A}} (X \cap Y)\cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y) \cup ( \bigcup^{\mathfrak{A}} (X\setminus Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y) = (\bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}}Y) \cup^{\mathfrak{A}} 0 = \bigcup^{\mathfrak{A}} (X \cap Y)\cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y.

Applying the formula \bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y twice we get

\bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup (Y \cap (X \cap Y)) = \bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} (X \cap Y) = \bigcup^{\mathfrak{A}} (X \cap Y).

But for any A,B\in R exist X,Y\in\mathscr{P}S such that A=\bigcup^{\mathfrak{A}}X and B=\bigcup^{\mathfrak{A}}Y. So A\cap^{\mathfrak{A}}B = \bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} (X \cap Y) \in R.

]]> <![CDATA[Complete lattice generated by a partitioning of a lattice element]]> http://portonmath.wordpress.com/2009/10/20/complete-lattice-generated-by-a-partitioning-of-a-lattice-element/ Mon, 19 Oct 2009 23:02:25 +0000 porton http://portonmath.wordpress.com/2009/10/20/complete-lattice-generated-by-a-partitioning-of-a-lattice-element/ In this post I defined strong partitioning of an element of a complete lattice. For me it was seeming obvious that the complete lattice generated by the set S where S is a strong partitioning is equal to \left\{ \bigcup{}^{\mathfrak{A}}X | X\in\mathscr{P}S \right\}. But when I actually tried to write down the proof of this statement I found that it is not obvious to prove. So I present this to you as a conjecture:

Conjecture The complete lattice generated by a strong partitioning S of an element of a complete lattice \mathfrak{A} is equal to \left\{ \bigcup{}^{\mathfrak{A}}X | X\in\mathscr{P}S \right\}.

Proposition Provided that this conjecture is true, we can prove that the complete lattice [S] generated by a strong partitioning S of an element of a complete lattice is a complete atomic boolean lattice with the set of its atoms being S (Note: So [S] is completely distributive).

Proof Completeness of [S] is obvious. Let A\in[S]. Then exists X\in\mathscr{P}S such that A=\bigcup{}^{\mathfrak{A}}X. Let B=\bigcup{}^{\mathfrak{A}}(S\setminus X). Then B\in[S] and A\cap B=0. A\cup B=\bigcup{}^{\mathfrak{A}}S is the biggest element of [S]. So we have proved that [S] is a boolean lattice.

Now let prove that [S] is atomic with the set of atoms being S. Let z\in S and A\in[S]. If A\neq z then either A=0 or x\in X where A=\bigcup^{\mathfrak{A}}X, X\in\mathscr{P}S and x\neq z. Because S is a strong partitioning, \bigcup^{\mathfrak{A}}(X\setminus\{z\})\cap^{\mathfrak{A}}z=0 and \bigcup^{\mathfrak{A}}(X\setminus\{z\})\neq 0. So A=\bigcup^{\mathfrak{A}}X=\bigcup^{\mathfrak{A}}(X\setminus\{z\})\cup^{\mathfrak{A}}z\nsubseteq z.

Finally we will prove that elements of [S]\setminus S are not atoms. Let A\in[S]\setminus S and A\neq 0. Then A\supseteq x\cup^{\mathfrak{A}}y where x,y\in S and x\neq y. If A is an atom then A=x=y what is impossible. QED

The above conjecture as a step to solution to the original conjecture may also be considered for the polymath research problem. Or maybe we should research both these two problems in a single polymath set, as the solution of one of them may inspire the solution of the other of these two problems.

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