When is the closure of an intersection equal to the intersection of closures?

Question

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?

Answer 1

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.

Answer 2

$\newcommand{\cl}{\operatorname{cl}}$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 $N,M\in\mathscr{N}$.)

Answer 3

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.

Answer 4

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.

Answer 5

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. $\square$