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I can reason this out intuitively (it seems obvious), but I can't seem to formalize a proof.

Prove $\left(\bigcup_{\alpha\in J} A_{\alpha}\right)^c = \left(\bigcap_{\alpha\in J} A^c_\alpha\right)$

Any thoughts?

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Is this a homework problem? –  Thomas E. Jan 24 '13 at 20:03
    
Yes, i'm sorry. –  Peej Gerard Jan 24 '13 at 20:08
    
The notion is just throwing me off on where to begin. I can see its a short proof, but its not like the other set proofs that I'm used to and thus, I've hit a roadblock. –  Peej Gerard Jan 24 '13 at 20:09
    
It's fine, you don't need to be. Just add the homework tag so people know what types of answers they should post. Also, it's always nice if you provide something that you have tried, so we know which part you can't get through. –  Thomas E. Jan 24 '13 at 20:10
    
thanks, i appreciate it –  Peej Gerard Jan 24 '13 at 20:10

2 Answers 2

up vote 1 down vote accepted

If $A$ and $B$ are two sets and you want to prove that $A=B$, then the standard way to do this is to show that $A\subset B$ and $B\subset A$. In order to show $A\subset B$, you need to show that $x\in A$ implies $x\in B$. Similarly for $B\subset A$ you need to show that $x\in B$ implies $x\in A$. I will help you to get started with the first implication.

Start by choosing $x\in (\bigcup_{\alpha\in J} A_{\alpha})^{c}$ and try to conclude that $x\in\bigcap_{\alpha\in J}A_{\alpha}^{c}$. Since $x\in (\bigcup_{\alpha\in J} A_{\alpha})^{c}$ then $x\notin \bigcup_{\alpha\in J} A_{\alpha}$ by definition of complement, so $x\notin A_{\alpha}$ for every $\alpha\in J$. Because if on the contrary there would exist $\alpha\in J$ so that $x\in A_{\alpha}$, then what would this say about the assumption $x\notin \bigcup_{\alpha\in J}A_{\alpha}$? Can you continue from here?

Remember that in general $x\in \bigcup_{i}A_{i}$ means that there exists $i$ so that $x\in A_{i}$ and $x\in\bigcap_{i}A_{i}$ means that $x\in A_{i}$ for all $i$.

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I follow your conclusion that $x\notin A_{\alpha}$ for every $\alpha\in J$. Isn't this equivalent to right hand side, $x\in\bigcap_{\alpha\in J}A_{\alpha}^{c}$? If $x\notin A_{\alpha}$ for every $\alpha\in J$, that is the same as, $x\in A^{c}_{\alpha}$ for every $\alpha\in J$. Which is what the righthand side says. –  Peej Gerard Jan 24 '13 at 20:33
    
Exactly. Can you show the other implication? –  Thomas E. Jan 24 '13 at 20:34
    
I assume we just reverse it. Sorry to be lazy with latex, but essentially we prove the subset in the other direction? –  Peej Gerard Jan 24 '13 at 20:38
    
If $x\in\bigcap_{\alpha\in J}A_{\alpha}^{c}$, then $x$ is in the intersection of all the sets $A^{c}_{\alpha}$. Can this imply that (applying the complement), $x$ is not the union of $A_{\alpha}$? This statement would seem to imply the lefthand side. –  Peej Gerard Jan 24 '13 at 20:41
    
True. You got it. –  Thomas E. Jan 24 '13 at 20:44

Hint: Recall that the definition of $x\in\bigcup_{i\in I} X_i$ is that for some $i \in I$, $x\in X_i$. Use the similar definition for intersection and complement and show two-sided inclusions.

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So if I let $x\in\bigcup_{\alpha\in J} A_\alpha$, that means that for some $\alpha \in J$, there exists an $x \in A_\alpha$? What does the complement of this entail? –  Peej Gerard Jan 24 '13 at 20:15
    
It means that $x$ is not in the union. What is the negation of "there exists ..."? –  Asaf Karagila Jan 24 '13 at 20:32
    
There does not exist...I think I see, so it essentially says that $x$ is in the intersection of the sets $A^{c}_{\alpha}$. Seems like all that was done was apply the definition of complements? –  Peej Gerard Jan 24 '13 at 20:37
    
Give and take. :-) –  Asaf Karagila Jan 24 '13 at 20:38
    
haha i'm sorry, you're saying that I'm correct, that is all we did essentially? –  Peej Gerard Jan 24 '13 at 20:42

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