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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?

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Assuming that $C(A)$ means the complement of $A$, then math.stackexchange.com/questions/207570/infinite-demorgan-laws is a duplicate. –  Asaf Karagila Jul 12 '13 at 19:34
    
Saying that $x\notin\bigcap_{\alpha\in I}A_\alpha$ means that $x\notin A_\alpha$ for at least an $\alpha\in I$, which is exactly $x\in\bigcup_{\alpha\in I}C(A_\alpha)$. –  egreg Jul 12 '13 at 19:34
    
@egreg What about the terms in $A_\alpha$ which are not in $\cap_{\alpha\in I}A_\alpha$? $x$ could be among those. –  Emrakul Jul 12 '13 at 19:39
    
@EmrakultheAeonsTorn Sorry, but I don't get it. –  egreg Jul 12 '13 at 19:55
    
@egreg There could be terns in $A_\alpha$ which are not in $\cap_{\alpha\in I}A_\alpha$. $x$ could still be in $A_\alpha$ even if $x\notin\cap_{\alpha\in I}A_\alpha$. –  Emrakul Jul 12 '13 at 19:57

1 Answer 1

up vote 2 down vote accepted

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 $$x\in \bigcap C(A_\alpha)$$

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