Sets Question [Showing Proof] I must show that for sets $A$, $B$, $C$, and the universal set $U$, that the following is true.
$U = A \cup B$ and $A \cap B = \emptyset$, then  $A = U\setminus B$.
I think this is true, but I don't know how to show a proof of it.
This is what I have:
Take $x$ in $A$.  Then $x$ is in $U = A \cup B$.  So $x$ is in $A$ or $B$.  But $A \cap B = \emptyset$, and $x$ is in $A$, so $x$ is not in $B$. Since $x$ is in $U$ and $x$ is not in $B$, $x$ is in $U\setminus B$.  So $A$ is contained in $U \setminus B$. For the reverse, take any $x$ in $U\setminus B$.  $x$ is in $U = A \cup B$.  $x$ is in $A$ or $B$.  $x$ is not in $B$ because $x$ is in $U \setminus B$,  so $x$ is in $A$.

Hint: At first, prove that $A \subset U \backslash B$. Then prove that $U \backslash B \subset A$. This is a typical way to prove two sets is equal.
As you require proof, for the first, let $x \in A$, then because $A \cap B = \varnothing$, one has $x \notin B$, hence $x \in U \backslash B$. For the second, similarly, but I think that you should do it by yourselves.
