Prove $A \cup A' = U$ and$ A \cap A' = \emptyset$ Prove $A \cup A' = U$ and $A \cap A' = \emptyset$
$A \cup A' = U$ 
set union definition with negation on the second $A$.
$[x: x \in A \lor x \notin A]$
This means that x is in $A$ or x is not in $A$. I wrote that since this is an or statement we can choose one or the other, but apparently that isn't a good reason to demonstrate that $A \cup A' = U$ . So, how do I prove it without set union definition? I don't think substituting $A \cup A' = U$  as $U=U$ is a good idea. That's too easy, and if that's the case, then I know it's wrong. 
$A  \cap A' = \emptyset$
is really easy to prove.
Using set definition of $A \cap A'$
$[x: x \in A \land x \notin A ]$
which tells me that x is in A and x is not an A. This is a very absurd statement. That's why it's an empty set. 
 A: Here is a calculational way to do these two proofs: expand definitions, simplify, and see where that leads you.
I'm assuming that $\;U\;$ is the universe we're working in, which contains every element, and that $\;{}'\;$ is set complement within $\;U\;$.
For every $\;x\;$ we have
\begin{align}
& x \in A \cup A' \\
\equiv & \qquad \text{"definition of $\;\cup\;$"} \\
& x \in A \lor x \in A' \\
\equiv & \qquad \text{"definition of $\;{}'\;$"} \\
& x \in A \lor \lnot (x \in A) \\
\equiv & \qquad \text{"logic: excluded middle"} \\
& \text{true} \\
\equiv & \qquad \text{"$\;U\;$ is our universe"} \\
& x \in U \\
\end{align}
Therefore, by set extensionality, $\;A \cup A' = U\;$.
In a very similar way we can prove $\;A \cap A' = \emptyset\;$: for every $\;x\;$
\begin{align}
& x \in A \cap A' \\
\equiv & \qquad \text{"definition of $\;\cap\;$"} \\
& x \in A \land x \in A' \\
\equiv & \qquad \text{"definition of $\;{}'\;$"} \\
& x \in A \land \lnot (x \in A) \\
\equiv & \qquad \text{"logic: contradiction"} \\
& \text{false} \\
\equiv & \qquad \text{"definition of $\;\emptyset\;$"} \\
& x \in \emptyset \\
\end{align}
A: Define $A' = \left\{ x : x \in U \text{ and } x \not \in A \right\} $. If $x \in A \cup A'$, then $x \in A$ or $x \in A'$, and either way $x \in U$, so $A \cup A' \subseteq U$.  If $x \in U$, then either $x \in A$ or $x \not \in A$ and so $x \in A \cup A'$, so that $U \subseteq A \cup A'$. Then, $U = A \cup A'$.
The second statement is only vacuously true where $A = \emptyset$, however you could also prove it via contradiction.
A: You must use set union definition. For every $x \in U$ we have that either $x \in A$ or $x \notin A$, that is, either $x \in A$ or $x \in A'$. Therefore $$x \in U \Rightarrow x \in A \cup A' \ .$$ The other way of inclusion is trivial since $A \subset U, \ A' \subset U$. Conclude that $$U \subset A' \cup A \subset U$$ and hence $U = A \cup A'$.
A: a) show that $A \cup A' \subseteq U $
$$ \forall_{ABC} A \subseteq C \land B \subseteq C \Rightarrow A \cup B \subseteq C  $$
trivial one. Therefore in particular $$ A \subseteq U \land A' \subseteq U \Rightarrow A \cup A' \subseteq U  $$
b) show that $U \subseteq A \cup A'$
$$\forall_x \; x \in A \lor x \notin A $$
$$_{(\text{this is $p \lor \neg p$ tautology})}$$
by definition $x \notin A \iff x \in A'$ therefore
$$\forall_x \; x \in A \lor x \in A' $$
so now we just
$$x \in U \Rightarrow x \in A \lor x \in A' \Rightarrow x \in A \cup A'$$
conclusion
a) and b) implicate that $A \cup A' = U $
now with $A \cap A'= \emptyset $
we will use the fact that
$$A=B \Rightarrow A'=B'$$
of course $$U'=\emptyset$$
and $$(A \cup A')'=A' \cap A = A \cap A'$$
if the first equality is not obvious to you then I'll show that
$$\forall_{AB} (A \cup B)'=A' \cap B'$$
here we go
$$x \in (A \cup B)' \iff x \notin A \cup B \iff x \notin A \land x \notin B \iff$$$$\iff x \in A' \land x \in B' \iff x \in A' \cap B'$$
$$$$
step $x \notin A \cup B \iff x \notin A \land x \notin B$ is taken from negating on both sides
$$x \in A \cup B \iff x \in A \land x \in B$$
$$$$
$$\neg (x \in A \cup B) \iff \neg(x \in A \land x \in B)$$
$$\neg (x \in A \cup B) \iff \neg(x \in A) \lor \neg(x \in B)$$
$$x \notin A \cup B \iff x \notin A \land x \notin B$$
