We know $\overline{\bigcap A_{\alpha}}\subseteq\bigcap\overline{A}_{\alpha} $, but when is the reverse inclusion true? Can you give some properties of the underlying space that would guarantee this?

  • 10
    $\begingroup$ This isn't true in a "nice" space such as $\Bbb{R}$. Consider the intersection of open intervals $(0,1/n)$ for natural numbers $n$. The intersection is empty (so closed), but the intersection of the closures is $\{0\}$. $\endgroup$
    – hardmath
    Apr 10, 2013 at 4:50
  • $\begingroup$ Maybe it holds for a compact metric space? $\endgroup$ Apr 10, 2013 at 4:54
  • 1
    $\begingroup$ If you like, put my example in the unit interval. $\endgroup$
    – hardmath
    Apr 10, 2013 at 4:55
  • 11
    $\begingroup$ Another good counterexample in $\mathbb{R}$: $(0, \frac{1}{2}) \cap (\frac{1}{2}, 1) = \varnothing$. $\endgroup$ Apr 10, 2013 at 5:07
  • $\begingroup$ One interesting case the reverse inclusion holds is if $A_\alpha$ are convex sets. See "Convex Analysis " R. Tyrrel Rockafellar, Theorem 6.5. $\endgroup$ Jul 27, 2023 at 5:53

5 Answers 5


This extends the answers of hardmath and Brian M. Scott, but completely answers the question.

Spaces satisfying the (seemingly) weaker condition that $\overline{ A \cap B } = \overline{A} \cap \overline{B}$ for all $A , B \subseteq X$ are discrete.

If $A \subseteq X$ is a non-closed set, pick $x \in \overline{A} \setminus A$. Note, then, that $$x \in \overline{ \{ x \} } \cap \overline{A} = \overline{ \{ x \} \cap A } = \overline{ \varnothing } = \varnothing,$$ which is absurd! Therefore all subsets are closed.

As hardmath noted, all discrete spaces satisfy the stronger condition in the OP, and so we have an equivalence of all three notions.

Given a topological space $X$, the following are equivalent:

  1. $\overline{ \bigcap_{i \in I} A_i } = \bigcap_{i \in I} \overline{A_i}$ for all families $\{ A_i \}_{i \in I}$ of subsets of $X$.
  2. $\overline{ A \cap B } = \overline{A} \cap \overline{B}$ for all $A , B \subseteq X$.
  3. $X$ is discrete.
  • $\begingroup$ Compare the equivalence of points (1) and (2) in this wonderful answer with the fact that, in any topological space $(X , \tau)$ the closure of the union of any two subsets of X equals the union of their closures, while it's not difficult to find a space $X$ and a family of subsets $(A_{i})_{i \in I}$ of it for which $\bigcup_{i \in I} \overline{A_{i }} \subsetneq \overline{\bigcup_{i \in I}A_{i}}$ ... $\endgroup$
    – Amelian
    Jan 16, 2022 at 18:58

$\newcommand{\cl}{\operatorname{cl}}$There is a very large class of spaces in which it fails. Let $X$ be any space with a non-isolated point $p$ such that the intersection of all nbhds of $p$ is $\{p\}$. Let $\mathscr{N}$ be the set of nbhds of $p$, and for each $N\in\mathscr{N}$ let $N'=N\setminus\{p\}$. Then

$$\bigcap_{N\in\mathscr{N}}\cl N'\supseteq\{p\}\ne\varnothing=\cl\bigcap_{N\in\mathscr{N}}N'\;.$$

This class includes all non-discrete $T_1$ spaces. (In case it isn’t immediately obvious that $p\in\cl N'$ for each $N\in\mathscr{N}$, note that $\mathscr{N}$ is a filter, and $\{p\}\notin\mathscr{N}$, so $N'\cap M\ne\varnothing$ for each $N,M\in\mathscr{N}$.)


It's trivially so if the underlying space has a discrete topology.

In such cases $\overline{A} = A$ for any subset $A$ of the space.

  • $\begingroup$ @hermes: Perhaps you overlooked my mention of the discrete topology? In which case every "collection of points is closed." $\endgroup$
    – hardmath
    May 27, 2020 at 19:40

Inspired by this answer, which showed that $A^\circ\cup B^\circ$ iff $\partial A\cap\partial B\subseteq\partial(A\cup B)$, here $A^\circ$ and $\partial A$ denoting respectively the interior and the boundary of $A$, I put a similar property on the two sets $A,B$ guaranteeing $\bar A\cap\bar B=\overline{A\cap B}$.

Noting that $\overline{A^c}=A^{c\circ}$ (abbreviation for $(A^c)^\circ$) and $\partial A=\partial(A^c)$, we have $$\begin{aligned}\overline{A\cap B}=\bar A\cap\bar B &\Leftrightarrow(A\cap B)^{c\circ}=A^{c\circ}\cup B^{c\circ}\\ &\Leftrightarrow(A^c\cup B^c)^\circ=A^{c\circ}\cup B^{c\circ}\\ &\Leftrightarrow\partial A^c\cap\partial B^c\subseteq\partial(A^c\cup B^c)\\ &\Leftrightarrow\partial A\cap\partial B\subseteq\partial(A\cap B). \end{aligned}$$ This is a necessary and sufficient condition for $\bar A\cap\bar B=\overline{A\cap B}$. As to the case of a family of sets, I don't know any similar condition.


We can also show that $X$ must be discrete by proving an interesting intermediate fact.

Proposition. Fix $B \subseteq X$. $\overline{A \cap B} = \overline A \cap \overline B$ for all $A \subseteq X$ iff $B$ is closed and open.

This holds in any topological space $X$. To prove it, we use

Lemma. Fix $U \subseteq X$. $\overline A \cap U \subseteq \overline{A \cap U}$ for all $A \subseteq X$ iff $U$ is open.

Proof of lemma. Here. $\square$

Proof of proposition.

$(\Rightarrow)$ Let $B=A^c$. Then $\varnothing = \overline \varnothing = \overline{A \cap B} = \overline A \cap \overline B$, and $X = \overline X = \overline{A \cup B} = \overline A \cup \overline B$. Therefore $B^c = A \subseteq \overline A = \big(\overline B\big)^c$. Taking complements, we have $\overline B \subseteq B$, hence $B$ is closed. Using this fact, we also have $\overline A \cap B = \overline A \cap \overline B = \overline{A \cap B}$, hence $B$ is open by the lemma.

$(\Leftarrow)$ Assume $B$ is closed and open. Then $\overline A \cap \overline B = \overline A \cap B \subseteq \overline{A \cap B}$, again using the lemma. $\square$


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