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I have a question regarding the Borel sequence of real sets. Given the collection $\cal{F}_\sigma$ as sets that can be expressed as the countable union of closed sets and $\cal{G}_\delta$ as the collection of sets expressable as the countable intersection of open sets, I need to prove that all open sets are of type $\cal{F}_\sigma$ and all closed sets are of type $\cal{G}_\delta$.

I need a hint on a how to proceed. I know the Lindelof property of open sets on the real line, that is, every open set is a countable union of open interval and every open interval is a countable intersection of closed intervals. Is this the correct way to proceed?

Edit I have corrected the mistake in my question regarding open and closed sets and the classes $\cal{F}_\sigma$ and $\cal{G}_\delta$.

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  • $\begingroup$ I edited to correct the definition of $\cal{G}_\delta$. By the way, are you sure you aren't asked to show open sets are $\cal F_\sigma$ and closed sets are $\cal G_\delta$? As the question stands, it's trivial. $\endgroup$ – Aweygan Sep 15 '18 at 16:51
  • $\begingroup$ You over-think it. Every closed set is a countable union of closed sets as $A=\bigcup A_i$ where all $A_i=A$. $\endgroup$ – Yanko Sep 15 '18 at 16:52
  • $\begingroup$ @Aweygan sorry my mistake I have corrected it. So now the hint? or an answer? $\endgroup$ – Iconoclast Sep 15 '18 at 20:31
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Let $A$ be a closed set. Since $\emptyset$ is a $\cal G_\delta,$ we may assume that $A\ne\emptyset.$ Then the function $$x\mapsto d(x,A)=\inf\{d(x,a):a\in A\}$$ is continuous, so for each $n\in\mathbb N$ the set $$U_n=\left\{x:d(x,A)\lt\frac1n\right\}$$ is open. I claim that $$A=\bigcap_{n\in\mathbb N}U_n.$$ The inclusion $A\subseteq\bigcap_{n\in\mathbb N}U_n$ is obvious. In the other direction, if $x\in\bigcap_{n\in\mathbb N}U_n,$ then $d(x,A)=0;$ since $A$ is closed, it follows that $x\in A.$

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  • $\begingroup$ So, your proof works for any Lindelöf space. $\endgroup$ – szw1710 Sep 15 '18 at 23:26
  • $\begingroup$ @szw1710 My proof is for metric spaces, whether Lindelöf or not. In a non-metrizable Lindelöf space, a closed set is not necessarily a $G_\delta.$ $\endgroup$ – bof Sep 15 '18 at 23:32
  • $\begingroup$ @szw1710 For example, let $X$ be an uncountable set, topologized so that the open sets are the empty set and the sets with countable complements. $X$ is a Lindelöf space, every countable set is closed, but no nonempty countable set is a $G_\delta.$ $\endgroup$ – bof Sep 15 '18 at 23:36
  • $\begingroup$ Thanks for the explanation. Of course your argument works in a metric space. $\endgroup$ – szw1710 Sep 16 '18 at 7:54
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Any open interval is a countable union of closed intervals, i.e is an $F_{\sigma}$ set. Indeed: $$ (a,b)=\bigcup_{n=n_0}^{\infty}\left[a+\frac{1}{n},b-\frac{1}{n}\right] $$ for $n_0$ large enough to create such intervals. Similarly for rhe intervals $[a,\infty)$ and $(-\infty,b]$. Then by the Lindelof property you mentioned, any open set, as a countable union of open intervals, is $F_{\sigma}$. The 2nd part follows immediately by complements, but it could be showwn directly by a similar way.

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