# Is it true that $A \cup (B - C) = (A \cup B) - (A \cup C)$?

I am brain farting on this question pretty hard, I'm asked to determine if

$$A \cup (B - C) = (A \cup B) - (A \cup C)$$

That is, determine if this set equality is true, and if not, which inclusions are true. I was able to show that equality fails by showing '$\subset$' does not hold using a relatively simple example, but I am having issues with opposite inclusion. My attempts began by assuming I had an arbitrary element $x \in (A \cup B) - (A \cup C)$ and tried to work my way down, but I haven't had much success.

Any help or hints would be appreciated, for I am too stubborn to move on to the next part without finishing this one.

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What do you mean by "which set inclusions hold"? – Christopher A. Wong Apr 1 '13 at 19:34
Whether $A \cup (B-C) \subset (A \cup B) - (A \cup C)$ or $(A \cup B) - (A \cup C) \subset A \cup (B-C)$ or both – Archie Apr 1 '13 at 19:34
OK, in that case, you should not have stated that those two expressions were equal. – Christopher A. Wong Apr 1 '13 at 19:42
@Christopher, Munkres, if I recall, gives an equality of sets, asks to prove or disprove it; then asks, if the equality doesn't hold, which inclusions, if any, hold. – amWhy Apr 1 '13 at 19:44
@amWhy Yes you are correct, I'll fix the phrasing. Sorry about that. – Archie Apr 1 '13 at 19:45

Suppose that $x\in(A\cup B)\setminus(A\cup C)$. Then by definition $x\in A\cup B$, and $x\notin A\cup C$. Since $x\notin A\cup C$, and $A\subseteq A\cup C$, you know that $x\notin A$. Thus, the only way to have $x\in A\cup B$ is to have $x\in B$. Now recall that $x\notin A\cup C$; this also implies that $x\notin C$, so $x\in B\setminus C$, and therefore $x\in A\cup(B\setminus C)$. Thus, $(A\cup B)\setminus(A\cup C)\subseteq A\cup(B\setminus C)$.
Hint: Think what happens when both $B$ and $C$ are disjoint from $A$. Are there any elements of $A$ in the RHS set? Are there any elements of $A$ in the LHS?