0
$\begingroup$

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.

  • 0
    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
  • 0
    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
  • 0
    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
  • 0
    do those 5 subsets make N5?2011-12-06
  • 0
    if yes, can that be proof that P(L) is not modular?2011-12-06
  • 1
    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
  • 0
    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

1 Answers 1

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 $$x\lor(r\land s)\ne(x\lor r)\land s=1\land s=s\;?$$
  • 0
    I really don't understand this :( can u write a solution, please, I need this for exam.2011-12-06
  • 0
    I don't know if power set and set of partitions and set of all subsets is the same thing, but P(L) is a set of all subsets of L.2011-12-06
  • 0
    @Ggg: I’ve expanded the hint considerably; see if that helps.2011-12-06
  • 0
    I think I can see it from Hasse diagram, right?2011-12-06
  • 0
    @Ggg: Yes, $P(L)$ is the set of all subsets of $L$, but $\{\{a,b,c,d,e\},\{f\}\}$, for instance, is not a subset of $L$, so it’s not a member of $P(L)$. It’s a *set* of subsets of $L$, so it’s a member of $P(P(L))$.2011-12-06
  • 0
    why it's not subset of L?2011-12-06
  • 0
    @Ggg: Because its elements are $\{a,b,c,d,e\}$ and $\{f\}$, which are not elements of $L$. They are *subsets* of $L$, but not *elements* of $L$.2011-12-06
  • 0
    I took a look at the problem - P(L) is a set of all partitions of L2011-12-06
  • 0
    @Ggg: That makes a big difference. It's important to know what everything means before even attempting a problem.2011-12-06
  • 0
    me and my bad english :) is now correct what I did at the start?2011-12-06