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Prove for every two sets $A$ and $B$ that $A-B$, ${B-A}$ and $A \cap B$ are pairwise disjoint.

I'm really stuck on this one. I know pairwise disjoint means no two elements in $A$ and $B$ are the same. $A \cap B$ is the empty set, but how would I prove this?

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    $\begingroup$ When talking about these three sets being pairwise disjoint, you need to show that $(A-B)\cap(B-A)=\varnothing$, $(A-B)\cap(A\cap B) = \varnothing$ and $(B-A)\cap (A\cap B) = \varnothing$. $\endgroup$ – John Habert Feb 28 '14 at 5:42
  • $\begingroup$ Note: It is not $A,B$ that are disjoint. $\endgroup$ – vadim123 Feb 28 '14 at 6:09
  • $\begingroup$ While writing out a nice verbal proof is good, for this, I think a Venn Diagram will make it clear what is going on. $\endgroup$ – Ryan Sullivant Feb 28 '14 at 6:28
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As you say two sets $A$ and $B$ are disjoint if $A \cap B = \emptyset$. But note that you want to show that $(A - B) \cap (B-A) = \emptyset$ and $(A - B) \cap (A \cap B) = \emptyset$ and $(B-A) \cap (A \cap B) = \emptyset$ as John points out in the comments.

First, can you see why $(A - B) \cap (B -A)$ should be empty? This is the set of all elements of $A$ that are not elements of $B$ while simultaneously being elements of $B$ but not elements of $A$. It would be weird if something satisfied both of these properties, right? Since we are trying to prove something about the emptyset, it may be best to do a proof by contradiction. So suppose $(A-B) \cap (B-A) \neq \emptyset$, i.e. there is some $x \in (A-B) \cap (B-A)$. Then $$x \in (A-B) \cap (B-A) \iff x \in (A - B) \wedge x \in (B - A)$$ $$\iff (x \in A \wedge x \notin B) \wedge (x \in B \wedge x \notin A)$$ $$\iff x \in A \wedge x \notin A \wedge x \in B \wedge x \notin B$$

So we have $x \in A \wedge x \notin A$, but this can definitely not happen with any element $x$; we have arrived at a contradiction, and our claim is false. Hence, $(A-B) \cap (B-A) = \emptyset$.

Now, for $(A-B) \cap (A \cap B)$, again try to think what this means. For something to be an element of this set it would have to be in $A$ but not $B$ but also in $A$ and $B$. This seems a bit fishy, huh?

We can proceed by contradiction like above, so suppose $(A-B) \cap (A \cap B) \neq \emptyset$ and let $x \in (A-B) \cap (A \cap B)$. Then, $$x \in (A-B) \cap (A \cap B) \iff (x \in A \wedge x \notin B) \wedge (x \in A \wedge x \in B) $$ $$\iff x \in A \wedge x \notin B \wedge x \in B$$ But now, we have $x \in B \wedge x \notin B$, again a contradiction. Thus, $(A-B) \cap (A \cap B) = \emptyset$.

I bet you can do $(B-A) \cap (A \cap B)$ since it is very similar to the previous one.

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