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 $latex S$ where $latex S$ is a strong partitioning is equal to $latex \left\{ \bigcup{}^{\mathfrak{A}}X | X\in\mathscr{P}S \right\}$. 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 $latex S$ of an element of a complete lattice $latex \mathfrak{A}$ is equal to $latex \left\{ \bigcup{}^{\mathfrak{A}}X | X\in\mathscr{P}S \right\}$.

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

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

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

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

1 thought on “Complete lattice generated by a partitioning of a lattice element

Leave a Reply

Your email address will not be published. Required fields are marked *