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

Based on the above, I want to prove that an arbitrary intersection of closed sets is closed.

Attempted proof: by De Morgan's theorem:

$(\bigcup_{G \in v}G)^{c} = \bigcap_{G \in v}G^{c} = B$. $B$ 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 my proof correct?

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    $\begingroup$ 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 $\endgroup$
    – kahen
    Jan 26, 2011 at 11:09
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    $\begingroup$ What's the question? $\endgroup$ Jan 26, 2011 at 11:23
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    $\begingroup$ 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. $\endgroup$ Jan 26, 2011 at 14:08
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    $\begingroup$ 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. $\endgroup$ Jan 26, 2011 at 15:51
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    $\begingroup$ Nitpick: $\{G^c|G\in v\}$ is not an arbitrary set of closed sets. Is a set of compliments of an arbitrary set of open sets. So this doesn't technically prove your claim. (Although it is trivially easy to fix: Let $v$ be an arbitrary collection of closed sets. Then $\cup_{H\in v} H^c$ is open and.....) $\endgroup$
    – fleablood
    Mar 15 at 6:25

1 Answer 1


This is true, and your reasoning is correct too.

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    $\begingroup$ This is a comment $\endgroup$
    – Heisenberg
    Jul 13, 2017 at 5:40
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    $\begingroup$ I have verified that the above comment is true. $\endgroup$ Jul 24, 2023 at 20:38

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