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October 20, 2009 / Victor Porton

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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