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October 20, 2009 / porton

Complete lattice generated by a partitioning – 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.

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