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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.

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    Do you mean for $r$ and $t$ to be **equivalence** relations, and for the underlying set to be $P(L\times L)$? Otherwise *block* makes no sense to me.2011-12-06
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    I don't understand your definition. "If $r$ and $t$ are relations"... relation on *what*? If they are relations on $L$, then they are elements of $P(L\times L)$, not of $P(L)$. If they are relations on $P(L)$, then they are not elements of $P(L)$. And what is a "block" of a relation?2011-12-06
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    No, I think it's not on LxL. I found all subsets of L, and I noticed r1={{a,b}, {c}, {d}, {e}, {f}} r2={{c,d,e}, {a}, {b}, {f}} r3={{a,b,c,f}, {d}, {e}} r4={{a,b,c,d,e}, {f}} r5={{a,b,c,d,e,f}}2011-12-06
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    do those 5 subsets make N5?2011-12-06
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    if yes, can that be proof that P(L) is not modular?2011-12-06
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    You’re looking at the lattice of partitions of $L$. The set of partitions of $L$ is not $P(L)$; it’s a subset of $P(P(L))$.2011-12-06
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    Your example doesn’t work. If it did, you’d expect the violation of modularity to be that $r_2\lor(r_3\land r_4)\ne (r_2\lor r_3)\land r_4$, but $$r_2\lor(r_3\land r_4)=r_2\lor \{\{a,b,c\},\{d\},\{e\},\{f\}\}=r_4=r_5\land r_4=(r_2\lor r_3)\land r_4\;.$$2011-12-06

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