In a metric space, it is true that every open set can be represented as the union of closed sets (e.g. the union of closed balls).

I am wondering if every open set can be represented as the countable union of closed sets?


2 Answers 2


Say $U$ is an open set. If $U$ is empty or if it is the whole space $X$ then $U$ is also closed, so we are done. Now assume that $U$ and its complement $U^c=X\setminus U$ are both non-empty.

Let $A_n = \bigcup_{x\in U^c} B(x,\dfrac1n)$. Notice that $A_n$ is open and its complement $A_n^c$ (which is closed) is contained in $U$. To finish off the problem show that $U$ is the union of the $A_n^c$.

A set which is the union of countably many closed sets is called an $F_\sigma$ set. It is known that every open set in a metric space is an $F_\sigma$ set.

Edit December 2023. I see I did not address the question whether every open set can be represented as the union of (any number of) closed sets. This is trivially true in every T1 space (i.e. when every point is closed). Every set is the union if its one-point subsets (singletons) and each singleton is closed.
$A=\cup\{\{x\}:x\in A\}$, and each singleton $\{x\}$ is closed.

  • $\begingroup$ How can we see $U \subseteq \bigcup_n A_n^c$? I know $A_n^c$ is an increasing sequence, is it enough? $\endgroup$
    – Messi Lio
    Dec 18, 2022 at 8:34
  • $\begingroup$ @MessiLio Not enough. But if $y\in U$ then since $U$ is open, there is some $n=n_y$ such that $B(y,\dfrac1n)\subseteq U.$ Then for any $x\in U^c$ we have $x\not\in B(y,\dfrac1n)$. Thus $y\not\in B(x,\dfrac1n)$, and $y\not\in A_n.$ Hence $y\in A_n^c.$ $\endgroup$
    – Mirko
    Dec 20, 2022 at 4:38

Let $U$ be an open subset of a metric space $E$. If $U\ne E$ then $$U=\bigcup_{n\in\mathbb N}\left\{x\in E:d(x,E\setminus U)\ge\frac1n\right\}$$ expresses $U$ as a countable union of closed sets.


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