Why is the intersection of finitely many $F_\sigma$ sets an $F_\sigma$ set, and the union of finitely many $G_\delta$ sets a $G_\delta$ set?
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4Please show you have put some effort into the question before asking it: be more specific about what you have tried and where you are stuck. – Ben Millwood Jan 30 '13 at 22:25
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Title and body seem to ask different questions. – Martin Jan 30 '13 at 22:43
1 Answers
First, the two statements are equivalent (by taking complements). Next, note that it suffices to prove that the union of two $G_\delta$-sets is a $G_\delta$-set, as the more general result then follows by induction.
Let $F=\bigcap_{n\in\Bbb N}H_n$ and $G=\bigcap_{n\in\Bbb N}K_n$ be $G_\delta$-sets in $X$, where each $H_n$ and $K_n$ is open. Then with repeated use of the distributive laws we have
$$\begin{align*} F\cup G&=\left(\bigcap_{n\in\Bbb N}H_n\right)\cup\left(\bigcap_{n\in\Bbb N}K_n\right)\\\\ &=\bigcap_{n\in\Bbb N}\left(H_n\cup\bigcap_{k\in\Bbb N}K_k\right)\\\\ &=\bigcap_{n\in\Bbb N}\left(\bigcap_{k\in\Bbb N}(H_n\cup K_k)\right)\\\\ &=\bigcap_{\langle n,k\rangle\in\Bbb N^2}(H_n\cup K_k)\;. \end{align*}$$
Each of the sets $H_n\cup K_k$ is open, and $\Bbb N^2$ is countable, so $F\cup G$ is the intersection of countably many open sets, i.e., a $G_\delta$-set.
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