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Show that any set contained in the metric space $(X, d)$ can be written as the intersection of open sets.

Definitions: A set $A \subseteq X$ is open if $\forall x \in A$, $\exists \varepsilon>0$ such that $B_{\varepsilon}(x) \subseteq A$. A set $C \subseteq X$ is closed if and only if $X \setminus C$ is open.

I also know that the intersection of a finite collection of open sets is open and that any open ball contained in $X$ is open. How can I prove this question?

José Carlos Santos
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1 Answers1

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Singletons $\{a\}$ are closed sets, so their complements are open. For any set $A \subseteq X$ consider the following: $$A=\bigcap_{a \in A^{c}}\{a\}^c.$$ Observe that each $\{a\}^c$ is open and once you verify the set equality (which is not that difficult) you have $A$ as the intersection of open sets.

influent
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Anurag A
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    Therefore the property relies not on $X$ being a metric space, but on it being a $T_1$ topological space – Maxime Ramzi Jul 24 '17 at 08:54
  • Thought: Would it work to say $$A = \bigcap_{n=1}^\infty \left{ x\in X \mid \mathrm{dist} (A, x) < \frac{1}{n} \right}$$ for an arbitrary $A\subseteq X$? If this set equality holds and the sets we intersect are really open, then we manage to solve this with a countable intersection. – Jeppe Stig Nielsen Jul 24 '17 at 13:18
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    @JeppeStigNielsen $X=\Bbb R$ and $A=X-{0}$ does not work with that trick (though it is even open by itself) – Hagen von Eitzen Jul 24 '17 at 13:30
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    @HagenvonEitzen You are right, thank you for that easy counterexample. (Addition: My method even fails for an open bounded interval inside $X=\mathbb{R}$.) I really felt like using the distance there. Is it true that any subset of a metric space is a countable intersection of open sets? – Jeppe Stig Nielsen Jul 24 '17 at 13:41
  • @JeppeStigNielsen I considered this worth a separate question – Hagen von Eitzen Jul 24 '17 at 13:57
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    @JeppeStigNielsen In a uniform space, the intersection of all uniform neighbourhoods of a set is the closure of the set. For metric spaces, we can do with a countable intersection and have $$\bigcap_{n = 1}^{\infty} \biggl{ x \in X : \operatorname{dist}(A,x) < \frac{1}{n}\biggr} = \overline{A}.$$ – Daniel Fischer Jul 24 '17 at 14:15
  • @DanielFischer Yeah, I was starting to realize something like that. I found a question with the problem to prove that every closed subset of a metric space was a countable intersection of open sets, and its answers used my $\bigcap_{n=1}^\infty \left{ x\in X \mid \mathrm{dist} (A, x) < \frac{1}{n} \right}$ approach. So since that is the thing that works in metric spaces, we could also ask about a "classification" of topological spaces with the property that every closed subset is the countable intersection of open subsets. – Jeppe Stig Nielsen Jul 24 '17 at 14:22
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    @JeppeStigNielsen https://en.wikipedia.org/wiki/Gδ_space – Adayah Jul 24 '17 at 16:06
  • I am not clear why the expression $A=\bigcap_{a \in A^{c}}{a}^c$ with respect to metric space X is valid for infinite intersection of open sets. – symmetric Aug 31 '23 at 14:44