Proof of statement I want to prove this statement:
$$(A_1 \cup A_2)^c  =  {A_1}^c \cap {A_2}^c$$
What I have realized so far, is that
$$(A_1 \cup A_2)^c \implies x \not\in A_1 \textrm{ and } x \not\in A_2 \implies x \not\in (A_1^c \cap A_2^c)$$ (Though I'm not sure if my approach is correct since I'm stuck)
but I am not sure how to proceed. Any help would be greatly appreciated.
 A: $x \in (A_1 \cup A_2)^c$ iff $x \notin A_1 \cup A_2$ iff $x \notin A_1 $ and $x \notin A_2$ iff $x \in A_1^c $ and $x \in A_2^c$ iff $x \in {A_1}^c \cap {A_2}^c$.
Hence $(A_1 \cup A_2)^c = {A_1}^c \cap {A_2}^c$.
A: First write down the definition of the first set, that is: 
for $\forall x \in (A_1 \cup A_2)^c$, $x$ is not in $A_1$ and/or $A_2$
$x$ is not in $A_1$ and $x$ is not in $A_2$
hence $x \in A_1^c \cap A_2^c$
hence $(A_1 \cup A_2)^c \subset A_1^c \cap A_2^c$
You can prove it also the other way around to get the equality , try it yourself.
A: This much of what you posted is correct:
$$(A_1 \cup A_2)^c \implies x \not\in A_1 \textrm{ and } x \not\in A_2$$
Continuing on, using first the definition of the complement of a set, and then the definition of set intersection: $$\begin{align} \underbrace{x \not\in A_1}_{\large\iff x\in A_1^c} \textrm{ and } \underbrace{x \not\in A_2}_{\large \iff x \in A_2^c} & \iff x \in A_1^c \underbrace{\text{ and }}_{\cap} x\in A_2^c \\ \\ &\iff x \in A_1^c \cap A_2^c\end{align}$$
