# How $\operatorname{cl}(A) \cap \operatorname{cl}(B) = \operatorname{cl}(A \cap B) \implies$ $X$ is the discrete space?

I know that if $$A,B \subset X$$ then $$\operatorname{cl}(A \cap B) \subset \operatorname{cl}(A) \cap \operatorname{cl}(B)$$ for every topological space $$X$$, but how I use that $$\operatorname{cl}(A) \cap \operatorname{cl}(B) \subset \operatorname{cl}(A \cap B)$$ to arrives that $$X$$ is the discrete topological space?

Suppose $$A \subseteq X$$, and note that our hypothesis (by which I assume you mean that for every $$A, B \subseteq X$$ we have $$\operatorname{cl}(A \cap B) = \operatorname{cl}(A) \cap \operatorname{cl}(B)$$) implies $$\emptyset = \operatorname{cl}(\emptyset) = \operatorname{cl}(A \cap A^\mathsf{c}) = \operatorname{cl}(A) \cap \operatorname{cl}(A^\mathsf{c})$$ and hence shows that $$\operatorname{cl}(A)$$ and $$\operatorname{cl}(A^\mathsf{c})$$ are disjoint.
On the other hand, in general we have $$X = \operatorname{cl}(X) = \operatorname{cl}(A \cup A^\mathsf{c}) = \operatorname{cl}(A) \cup \operatorname{cl}(A^\mathsf{c})$$ so we find that $$\operatorname{cl}(A^\mathsf{c}) = \operatorname{cl}(A)^\mathsf{c}$$.
But now consider the fact that $$A \subseteq \operatorname{cl}(A)$$ and $$A^\mathsf{c} \subseteq \operatorname{cl}(A^\mathsf{c}) = \operatorname{cl}(A)^\mathsf{c}$$. The latter inclusion implies $$\operatorname{cl}(A) \subseteq A$$, so combined with the former inclusion we see that $$\operatorname{cl}(A) = A$$.
In other words, we have shown that an arbitrary set is closed, and hence $$X$$ is discrete.