# A question on closure of intersection of open sets being equal to the intersection of closure of open sets

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?

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, $$\overline{(-1,0)}\cap\overline{(0,1)}=\{0\}.$$
• 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. Nov 20, 2022 at 4:08