Boolean algebra-Modular lattice

Question

Let $L=\{a,b,c,d,e,f\}$; $P(L)$ is the set of all partitions of $L$, and $\le$ is the order relation on $P(L)$ defined as:

if $r$ and $t$ are relations, then $r\le t$ iff every block in $r$ is a subset of some block in $t$.

Show that the lattice $(P(L),\le)$ is not modular.

Answer 1

Your comments indicate that you’re really looking at the lattice $(P,\le)$ of partitions of $L$, where for $r,s\in P$ we define $r\le s$ iff for each $x\in r$ there is a $y\in s$ such that $x\subseteq y$. (That is, each piece of $r$ is a subset of some piece of $s$.) Note that $P$ is not $\wp(L)$, or even a subset of $\wp(L)$: it’s a subset of $\wp(\wp(L))$.

HINT: Let $1$ be the trivial partition whose only member is $\{a,b,c,d,e,f\}$, and let $0$ be the partition $\{\{a\},\{b\},\{c\},\{d\},\{e\},\{f\}\}$. Let $$r=\{\{a,b,c\},\{d,e,f\}\}$$ and $$s=\{\{a,d\},\{b,e\},\{c,f\}\}\;.$$

1. Can you show that $r\land s=0$ and $r\lor s=1$? That is, can you show that $0$ is the only partition $\le$ both $r$ and $s$, and $1$ is the only partition $\ge$ both $r$ and $s$?
2. Can you find a partition $x$ of $L$ such that $x=x\le s$ and $$x\lor(r\land s)\ne(x\lor r)\land s=1\land s=s\;?$$