I am quite intrigued by this problem mainly because the result holds if the sets $A$ and $B$ are open, but why is openness required and how is it relevant to the equality of both sets?

Suppose $X$ is a topological space endowed with a topology $\mathscr{T}$. If $A, B \in \mathscr{T}$, show that $\overline{A \cap B} = \overline{A} \cap \overline{B}$.

Well, it is fairly easy to prove that $\overline{A \cap B} \subseteq \overline{A} \cap \overline{B}$. This is because $A \subseteq \overline{A}$ and $B \subseteq \overline{B}$ which gives $A \cap B \subset \overline{A} \cap \overline{B}$ and thus, since $\overline{A} \cap \overline{B}$ is closed, we have $\overline{A \cap B} \subseteq \overline{A} \cap \overline{B}$ since $\overline{A \cap B}$ is the smallest closed set containing $A \cap B$.

The other direction of inclusion is what bugs me. Well, I know that $A$ and $B$ must be open subsets of $X$ but apart from that, how do I proceed? How would I be able to relate the openness of $A$ and $B$ to the desired conclusion? I have read somewhere (correct me if I am wrong) that if $U$ and $V$ are open, then $\overline{U \cap V} = \overline{U \cap \overline{V}}$ but how is this relevant to the desired conclusion?


1 Answer 1


I don't believe this result holds in general. For example, consider $\mathbb{R}$ with the standard topology. Then $(-1,0)$ and $(0,1)$ are open sets with $$\overline{(-1,0)\cap(0,1)}=\overline{\emptyset}=\emptyset.$$ However, $$\overline{(-1,0)}\cap\overline{(0,1)}=\{0\}.$$

  • $\begingroup$ There is a result here (which does not have any proof why it holds) that seem to support the claim (closure and intersection commute). Although I don't know which assumptions am I missing for the result to hold. Link: en.wikipedia.org/wiki/Interior_(topology) $\endgroup$ Nov 20, 2022 at 3:56
  • 1
    $\begingroup$ If you are talking about the theorem listed below the heading "Interior Operator" that theorem says the interior of an intersection of closures equals the interior of the closure of the intersections in a complete metric space $X$. That is different than your claim. $\endgroup$
    – badatalg
    Nov 20, 2022 at 4:08

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