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?

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    $\begingroup$ 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 '13 at 4:50
  • $\begingroup$ Maybe it holds for a compact metric space? $\endgroup$ – Forever Mozart Apr 10 '13 at 4:54
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    $\begingroup$ If you like, put my example in the unit interval. $\endgroup$ – hardmath Apr 10 '13 at 4:55
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    $\begingroup$ Another good counterexample in $\mathbb{R}$: $(0, \frac{1}{2}) \cap (\frac{1}{2}, 1) = \varnothing$. $\endgroup$ – manthanomen Apr 10 '13 at 5:07

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.

$\newcommand{\cl}{\operatorname{cl}}$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.

  • $\begingroup$ This is not true. For each point is closed, but by the fact that finite union of closed sets is closed, only finite collection of points is closed. So infinite collection of points $A$ is open, which is also a Frechet filter. Thus $\overline{A}\neq A$ $\endgroup$ – hermes May 27 '20 at 16:33
  • $\begingroup$ @hermes: Perhaps you overlooked my mention of the discrete topology? In which case every "collection of points is closed." $\endgroup$ – hardmath May 27 '20 at 19:40
  • $\begingroup$ But it would make no sense (or trivial) because there would be no open sets of (points) or every set is both open and closed $\endgroup$ – hermes May 27 '20 at 19:43

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