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Prove that every subset of a metric space $M$ can be written as the intersection of open sets.

My attempt:

If $A\subset M$ is open, $A$ can be written as $A\cap M$, which is the intersection of 2 open sets.

If $A\subset M$ is closed, it can presumably be written as the intersection of infinitely many open sets. But I'm not sure how.

And what if $A\subset M$ is neither open nor closed?

Siddhartha
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  • Hint: can you show that any single point is an intersection of a collection of open sets? Once you can do this, you can achieve any other set using the unions of the open sets you used for each of its points. – Chessanator Sep 14 '18 at 17:16
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    It might be easier to find the proof if you consider the equivalent statement that every subset is a union of closed subsets. – Andreas Blass Sep 14 '18 at 17:19

2 Answers2

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Given $A$, let $B$ be the intersection of all open sets containing $A$. Then show $A=B$: Clearly $A\subseteq B$. Can you see why $x\notin A$ implies $x\notin B$?

Hagen von Eitzen
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Every singleton is closed in a metric space. That's all we need. (Simple proof of this fact, for any $p \in X$: $$X \setminus \{p\} = \bigcup \{B(q, d(p,q)): q \neq p\}$$ showing its complement is a union of open balls, hence open.)

Then observe that $$A = \bigcap \{X\setminus \{p\}: p \notin A\}$$

is an intersection of open sets for any $A \subseteq X$.

Henno Brandsma
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