Prove that $C\left(\bigcup_{\alpha\in I} A_\alpha\right)=\bigcap_{\alpha\in I}C\left(A_\alpha\right)$ I'm working through a book on fundamental topology, and I'm stuck at a proof in set theory. The book provides a proof for: 
$$\text{Let }\left\{A_\alpha\right\}_{\alpha\in I}\text{ be an indexed family of subsets of a set }S\text{; then:}$$
$$C\left(\bigcup_{\alpha\in I}A_\alpha\right)=\bigcap_{\alpha\in I}C\left(A_\alpha\right)$$
where $I$ is an indexing set for $\alpha$, and $\alpha\in I$, and $C(S)$ denotes the complement set. They do this by proving first that, if $x\in C\left(\bigcup_{\alpha\in I}A_\alpha\right)$, then $C\left(\bigcup_{\alpha\in I}A_\alpha\right)\subset\bigcap_{\alpha\in I}C\left(A_\alpha\right)$. Then, the prove the converse; that if $x\in\bigcap_{\alpha\in I}C\left(A_\alpha\right)$, then $\bigcap_{\alpha\in I}C\left(A_\alpha\right)\subset C\left(\bigcup_{\alpha\in I}A_\alpha\right)$, so the two sides, being subsets of each other, must be equal.
That makes perfect sense, and is very clear and straightforward. However, what I don't understand is their second proof (under the same conditions):
$$C\left(\bigcap_{\alpha\in I}A_\alpha\right)=\bigcup_{\alpha\in I}C\left(A_\alpha\right)$$
While this statement intuitively makes sense to me, I can't seem to prove it mathematically. Here's what I've done so far:
Let $x\in C\left(\bigcap_{\alpha\in I}A_\alpha\right)$. We know that, conversely, $x\notin \bigcap_{\alpha\in I}A_\alpha$. This is where the proofs diverge, though; if it were the first formula, it would be possible to say that $x\notin \bigcup_{\alpha\in I}A_\alpha$, and as a result, that $x\notin \left\{A_\beta\right\}_{\beta\in I}$. However, I can no longer do that, and approaching it from the other direction gives the same result.
Let $x\in\bigcup_{\alpha\in I}C\left(A_\alpha\right)$. We know that $x\notin\bigcup_{\alpha\in\ I}A_\alpha$, but this still doesn't get me anywhere. 
How should I approach this proof? Any hints or pointers for someone new to topology?
 A: Careful "Let $x\in\bigcup_{\alpha\in I}C\left(A_\alpha\right)$. We know that $x\notin\bigcup_{\alpha\in\ I}A_\alpha$". From $x\in\bigcup C(A_\alpha)$ you get that $x\in C(A_\alpha)$ for at least one $\alpha$. I give you the proofs below. Note how the quantifiers $\forall \alpha $ and $\exists \alpha$ are used, that is, "for all $\alpha$", "for each $\alpha$" and "there exists an $\alpha$", "for some $\alpha$".

What does it mean that $$x\in\bigcup C(A_\alpha)?$$
It means that $x\in C(A_\alpha)$ for at least one $\alpha$; which means $x\notin A_\alpha$ for at least one $\alpha$. But this means $x\notin \bigcap A_\alpha$, which means exactly that $$x\in C\left(\bigcap A_\alpha\right)$$
What dos it mean that $$x\in C\left(\bigcap A_\alpha\right)?$$
It means that $x\notin \bigcap A_\alpha$. This means that $x\notin A_\alpha$ for at least one $\alpha$, so $x\in C(A_\alpha)$ for at least one$\alpha$. But this means exactly $$x\in \bigcup C(A_\alpha)$$

What does it mean that $$x\in\bigcap C(A_\alpha)?$$
It means that $x\in C(A_\alpha)$ for each $\alpha$; which means $x\notin A_\alpha$ for each $\alpha$. But this means $x\notin \bigcup A_\alpha$, which means exactly that $$x\in C\left(\bigcup A_\alpha\right)$$
What dos it mean that $$x\in C\left(\bigcup A_\alpha\right)?$$
It means that $x\notin \bigcup A_\alpha$. This means that $x\notin A_\alpha$ for each $\alpha$, so $x\in C(A_\alpha)$ for each $\alpha$. But this means exactly $$x\in \bigcap C(A_\alpha)$$
