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It can be proved that arbitrary union of open sets is open. Suppose $v$ is a family of open sets. $\cup_{G \in v}G = A$ is an open set.

Now by De Morgan's theorem:

$(\cup_{G \in v}G)^{c} = \cap_{G \in v}G^{c} = B$. $B$ is a closed set since it is the complement of open set $A$.

$G$ is an open set, so $G^{c}$ is a closed set. $B$ is an infinite union intersection of closed sets $G^{c}$.

Hence Infinite intersection of closed sets is closed. Is it not a wrong conclusion.

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Slight pedantry: It can't be proven that arbitrary unions of open sets are open. That's part of the definition of open sets in a topological space –  kahen Jan 26 '11 at 11:09
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What's the question? –  Alexei Averchenko Jan 26 '11 at 11:23
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kahen: Well, that depends on your axioms. A topological space may as well be defined in terms of closed sets, or the closure/interior-operations. –  Fredrik Meyer Jan 26 '11 at 14:08
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To second Alexei's comment: what is the point of this post? Did you perhaps forget to include the question? And what's up with that last sentence? It sounds like you are defending yourself from criticism, but as far as I can tell no-one is saying that your conclusion is wrong. –  Willie Wong Jan 26 '11 at 15:51

1 Answer 1

up vote 5 down vote accepted

This is true, and your reasoning is correct too.

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