# Proving a relation between 2 sets as antisymmetric

Let $U = \{1,...,n\}$
And let $A$ and $B$ be partitions of the set $U$ such that:
$\bigcup A = \bigcup B = U$
and $|A|=s, |B|=t$
Let's define a relation between the sets $A$ and $B$ as follows:
$B \succ A \iff \forall_{1 \leq i \leq t}, \exists_{1 \leq j \leq s}: B_i \subseteq A_j$
Now, we want to prove that $\succ$ is an antisymmetric relation. Thus, we want to prove that:
$\forall_{A,B},A \succ B \land B \succ A \implies A=B$
How would one prove this? I'm pretty much stuck after laying out the conditions, and I don't have a clue on what assumptions I should make that will lead me anywhere?
PS: I hope I typed all that LaTeX up correctly. :)

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@AustinMohr: You mean: if $A\succ B$ and $A\ne B$, then $B\not\succ A$, right? –  Henning Makholm Aug 4 '12 at 22:09

First of all, for what you write to make sense, you're not picking two partitions and defining a relation between those particular two partitions. You're defining a single relation on the set of all partitions.

Second, I think you're overcomplicating things by considering a partition to be an indexed family of subsets of $U$. Indeed, if you consider the assignment of indices to be part of the partition, then what you want to prove is not true, because then the two partitions

$$A_1 = \{1\}, A_2 = \{2,3,\ldots,n\}$$ and $$B_1 = \{2,3,\ldots,n\}, B_2 = \{1\}$$ would satisfy $A\succ B\succ A$, but $A\ne B$.

So we need to work with partitions being unordered collections of subsets of $U$, and your relation should then be defined as $$B\succ A \quad\iff \forall b\in B \exists a\in A: b\subseteq a$$

In order to prove that this is antisymmetic, we assume $A\succ B\succ A$ and seek to prove that $B \subseteq A$. (Then, since $A$ and $B$ were arbitrary, and we also have $B\succ A\succ B$, the same argument shows $A\subseteq B$, so $A = B$).

In order to prove this, it is crucial that $A$ and $B$ are partitions, which requires among other things that (1) any two different members of $A$ must be disjoint, and (2) the empty set is not in $A$.

Now, we're assuming that $A\succ B\succ A$. To prove $B\subseteq A$ consider any $x\in B$. One of the $\succ$s gives us $y\in A$ such that $x\subseteq y$, and the other gives $z\in B$ sucht that $x\subseteq y\subseteq z$. But because $x$ and $z$ are both in $B$, they must be either equal or disjoint. Since $x$ is non-empty and $x\subseteq z$ they can't be disjoint, so $x=z$. But then $x\subseteq y \subseteq x$, and $y$ must equal $x$. Since $y$ was in $A$, we have proved $x\in A$.

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Just what I was looking for! Thanks! –  David D Aug 4 '12 at 22:50
$$\forall_{A,B},A \succ B \land B \succ A \implies A=B$$
We start with the quantifier; here it is $\forall_{A,B}$. So let's pick two arbitrary partitions $A$ and $B$.
Next we notice that we want to prove an implication. One proof strategy to use here is to assume the antecedant (the stuff on the left of the $\implies$) and try to prove the conclusion (the stuff on the right of the $\implies$).
Now we have reduced the problem to assuming that $A \succ B \land B \succ A$ and prove that $A = B$. How do you prove that two partitions are equal? I'll let you think about this for a little bit then come back and tell us what you come up with and where you get stuck.