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I doing some exercise related to $G_{\delta}$ set and got something confused. From the definition of topology space, finite intersection of finite open sets is an open set. By induction, we can conclude that countable intersection of open sets is open too (I see a lot of proof concluding that in set theory). But then $G_{\delta}$ notion is the same as open set, which I think impossible. Can some one clarify for me?

Thanks so much.

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    $\begingroup$ No: $$\bigcap_{n=1}^\infty \left(-1-\frac1n, 1+\frac1n\right) = [-1,1].$$ $\endgroup$
    – njguliyev
    Sep 5, 2013 at 18:01
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    $\begingroup$ Induction doesn't apply here. You can't go from "an intersection of $n$ open sets is open for all $n \in \mathbb{N}$" to "an intersection of countably many open sets is open". $\endgroup$ Sep 5, 2013 at 18:02

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You can't use induction like that. If you started with the fact that the intersection of two open sets is open, induction will get you that for any finite number of open sets, their intersection is open. But there is no way to go from that to a countable intersection of open sets is open. Which is good because, as seen in the other answers, that is false.

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In general no, just take intervals $(-1/n, 1/n)$ for $n=1,2,\ldots$ as open subsets of the real line. Then the countable intersection of all these is just the point $0$, which is not an open set.

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