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Let $X$ be a topological space and fix a sequence of closed sets $(C_n)$ such that $C_n\subseteq C_{n+1}$ for all $n$. Then $C:=\bigcup C_n$ is an $F_{\sigma}$-set. Fix also a set $S\subseteq C$. Is it true that $$ \mathrm{Closure}(S) \subseteq C\,\,\,? $$

My attempt: For each $n$ we have $C_n \supseteq C_n \cap S$. Hence $C_n \supseteq \mathrm{Closure}(C_n \cap S)$ hence $$ C = \bigcup C_n \supseteq \bigcup \mathrm{Closure}(C_n \cap S). $$ At this point, is it true that $\bigcup \mathrm{Closure}(C_n \cap S)=\mathrm{Closure}(S)$?

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  • $\begingroup$ I added an example with $S$ countable. $\endgroup$ – positrón0802 Jul 18 '17 at 22:45
  • $\begingroup$ I removed the edit because I already answered myself, but thanks anyway! $\endgroup$ – user207096 Jul 18 '17 at 22:50
  • $\begingroup$ (1). What if S=C and C is not closed in X? (2). What if S is any dense subset of the subspace C and C is not closed in X? $\endgroup$ – DanielWainfleet Jul 19 '17 at 17:13
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Let $X= \mathbb{R}$ in the usual topology. Enumerate the rationals as $r_n, n=1,2,\ldots$, and define $C_n = \{r_1, \ldots, r_n\}$ which are all finite thus closed. Then $C= \mathbb{Q}$ and $S= \mathbb{Q}$ has closure $\mathbb{R}$..

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Consider $X=\mathbf{R}$ in the standard topology. Let $C_n=[0,1-1/n]$ for each $n.$ Then $C=[0,1).$ If $S=C=[0,1)$ then $\overline{S}=[0,1] \not\subset C.$

In general, if $C$ is not closed you can choose $S=C$ and $\overline{S} \not\subset C.$

EDIT:

Now that you add the condition that $S$ must be countable.

Let $\mathbf{R}$ having all sets $(-\infty,a)=\{ x : x<a \}$ as basis. For each $n$ let $C_n=[1/n, +\infty).$ Then each $C_n$ is closed and $C=(0, +\infty).$ Consider $S=\{ 1/n : n \in \mathbf{Z}_+ \}.$ Then $S$ is countable and $S \subset C,$ but we can show that in this topology $\overline{S}=[0,+\infty),$ so $\overline{S}\not\subset C.$

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