True or false? $(X\setminus Y)\cup(Y\setminus Z)\cup(Z\setminus X) = X\cup Y\cup Z$, for any sets $X$, $Y$, $Z$. Question in my proofs homework, I am not too familiar with sets. How would I start with this? Maybe contradiction?
 A: If you cannot think of a counterexample you can just try proving the equation. If you succeed you have shown the claim to be true, and otherwise you might get clues for a counterexample.
Let us go through a simpler example: is $(X-Y)\cup(Y-X)=X\cup Y$? To prove two sets equal it is often easiest to show that they are contained in each other, so let's try this.
It should not be too difficult to prove that $(X-Y)\cup(Y-X)\subseteq X\cup Y$, and I'll leave that as an exercise. You start by assuming $x\in(X-Y)\cup(Y-X)$ and then show that $x\in X\cup Y$.
For the other direction, suppose $x\in X\cup Y$. Then either $x\in X$ or $x\in Y$. Suppose $x\in X$. If also $x\notin Y$, then we have $x\in X-Y$, and hence $x\in(X-Y)\cup(Y-X)$. But what if in addition to $x\in X$ we have $x\in Y$? Then $x\notin X-Y$, but also $x\notin Y-X$, so in fact $x\notin(X-Y)\cup(Y-X)$.
So, we got in trouble in the case when both $x\in X$ and $x\in Y$ hold. This suggests that a counterexample might be found by finding sets $X$ and $Y$ such that there is an element $x\in X\cup Y$ with both $x\in X$ and $x\in Y$.
A: Consider what happens to elements in all three sets. 
First, lets reword your theorem: 
If $X$, $Y$, and $Z$ are sets, then $a$ $\in$$X$ $\cup$ $Y$ $\cup$ $Z$ $\leftrightarrow$ $a$$\in$ ($X-Y$) $\cup$ ($Y-Z$) $\cup$ ($Z-X$).
Suppose that the theorem is true. Then we will show that the forward implication is a contradiction. Suppose $a$ $\in$ $X$ $\cap$ $Y$ $\cap$ $Z$, then $a$ satisfies the premise for the forward implication because ($X$ $\cap$ $Y$ $\cap$ $Z$) $\in$ ($X$ $\cup$ $Y$ $\cup$ $Z$). However, any element with this further property of intersection will not be in ($X-Y$) $\cup$ ($Y-Z$) $\cup$ ($Z-X$), since, for example, the element fails to be in $X-Y$ (or any of the others), since it already in all three!
I think Darrin had a good example listed in the comments for your question explaining this. Imagine a simple set with one element. The contradiction I have outlined becomes quite clear.
