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1) For me it is clear that a countable union of $F_\sigma$ sets is $F_\sigma$, but how to prove that countable INTERSECTION of $F_{\sigma}$ sets is $F_{\sigma}$ set ?
2) Prove that countable intersection of $G_{\delta}$ sets is $G_{\delta}$ set.

I tried to show (2):
$$l \in \mathbb{N}, A_{l}\textit{ is } G_{\delta}, A_{l}=\bigcap\limits_{n \in \mathbb{N}}V_{n}^{l}$$ $$\bigcap\limits_{l \in \mathbb{N}}A_{l}= \bigcap\limits_{l \in \mathbb{N}} \bigcap\limits_{n \in \mathbb{N}}V_{n}^{l} = \bigcap\limits_{i \in \mathbb{N}}W_{i}$$ $$ x \in \bigcap\limits_{i \in \mathbb{N}}W_{i}\Leftrightarrow \forall i\in \mathbb{N} \ x \in V_{n}^{l}\Leftrightarrow x \in \bigcap\limits_{n \in \mathbb{N}}\bigcap\limits_{l \in \mathbb{N}}V_{n}^{l} $$

Joe
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    Please, don't add questions after you've been answered. It makes existing answers look silly. –  Jan 08 '18 at 18:37
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    A countable intersection of $F_\sigma$ sets is called an $F_{\sigma\delta}$ set. – GEdgar Jan 09 '18 at 01:05

1 Answers1

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Claim (1) is not true.

$\Bbb R^n$ is a second-countable metric space, therefore open sets are $F_\sigma$. Your claim specialized to this case would then imply that all $G_\delta$ sets of $\Bbb R^n$ are $F_\sigma$. And, therefore, that $F_\sigma$ is a $\sigma$-algebra on $\Bbb R^n$. Specifically, $F_\sigma=G_\delta=\text{Borel}$. But this is not the case.

For (2) you can use the lemma for countable union of $F_\sigma$ and the fact that a set is $G_\delta$ if and only if it is the complement of a $F_\sigma$.

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    Open sets are $F_{\sigma}$ in any metric space....The Baire Category Theorem implies that a dense $G_{\delta}$ subset of $\Bbb R$ is uncountable, so $ \Bbb R$ \ $\Bbb Q$ is a $G_{\delta}$ set which is not $F_{\sigma}.$...................+1 – DanielWainfleet Jan 09 '18 at 14:44
  • @DanielWainfleet Ah, true. I could have used the distance from the complement. –  Jan 09 '18 at 14:49