# Proofs involving $(A\setminus B) \cup (A \cap B)$

Give short arguments for each part below to complete the reasoning. Please do not use Venn diagrams.

a. Prove that $A = (A\setminus B) \cup (A \cap B)$. (Hint: You should prove that $(A\setminus B)\cup (A\cap B) \subseteq A$ and vice-versa)

b. Prove that $(A \setminus B) \cap (A \cap B) = \emptyset$ (Hint: Try a proof by contradiction for this statement.).

I'm not too sure exactly where to start on this. Both statements make sense logically, but I don't know how exactly to prove them. Can anyone point me in the right direction?

(a). see here .
(b). If $x\in A \setminus B,$ then $x\notin B$. Hence $x\notin A\cap B$

Since $A\setminus B\subseteq A$ and $A\cap B\subseteq A$, we have $(A\setminus B)\cup(A\cap B)\subseteq A$.

Let $a\in A$. There are two cases: either $a\in B$ or $a\notin B$.

If $a\in B$, then $a\in A\cap B$; if $a\notin B$, then $a\in A\setminus B$. Hence $A\subseteq(A\setminus B)\cup(A\cap B)$.

Suppose $x\in(A\setminus B)\cap(A\cap B)$. Then $x\in A\setminus B$ implies $x\notin B$. On the other hand…

$x\in A\cap B$ implies $x\in B$.