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Show that every closed subset of a metric space is the intersection of a countable number of open sets.

This is Armstrong's Basic Topology Exercise 2.30 and it's following the section on Tietze extension theorem. I'm thinking about using the Tietze extension theorem, which states:

Any real-valued continuous function defined on a closed subset of a metric space can be extended over the whole space.

But I have no clue how to construct the open sets that we need to intersect...so maybe I'm not on the right track. Any hint please (to the question itself, not necessarily using Tietze extension theorem?)

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    $\begingroup$ Hint: You get an open set (why?) if you put tiny balls about each point of the set and take the union of these balls. Now suppose all the balls have the same radius, say $\frac{1}{n}.$ Now consider the intersection . . . $\endgroup$ – Dave L. Renfro Jan 7 '16 at 17:29
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Hint: For each $x\in C$, the closed set, consider the coverings $\mathcal{A_n}=\{B(x,1/n)\}$.

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  • $\begingroup$ I see. So $C \subseteq \cap A_n$ because for each $x \in C$, $x \in B(x, 1/n)$ for all $n$. Also, $\cap A_n \subseteq C$. Since if $p \notin C$, then there exists an $\epsilon > 0$ such that $\inf d(x, p) > \epsilon$ for all $x \in C$. Then we can take $1/n < \epsilon$, so $p \notin A_{1/\epsilon}$ therefore not in $\cap A_n$. How doe this look? $\endgroup$ – nekodesu Jan 7 '16 at 17:37
  • $\begingroup$ FYI (Shiyue), notice that we can't remove the assumption that $C$ is closed in your argument, as you can see with $C = \mathbb Q$ in the reals. Question to ponder (and how math research can start): Can "$C$ is closed" be replaced by a weaker assumption on $C$ (i.e. a property that all closed sets have, but a property that is also satisfied by some non-closed sets)? Maybe start by considering this for $C \subset {\mathbb R}^{n}.$ Note that the answer can be "NO" for some spaces, such as a discrete metric space, so the nontrivial question is whether it can be "YES" for some metric space. $\endgroup$ – Dave L. Renfro Jan 7 '16 at 17:54
  • $\begingroup$ As long as you say that $A_n=\cup_{x\in C} B(x, 1/n)$ beforehand it seems fine. Also I would say choose $N$ large enough so that $1/N<\epsilon$ and hence $p\notin A_N$ $\endgroup$ – Foobaz John Jan 7 '16 at 17:56

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