Let A and B be subsets of a topological space (X, τ). Prove that $\overline{A\cap B} \subseteq \overline{A} \cap \overline{B}$

Question

Hi I was working on the following problem :

I have some kind of intuition why this would be (tough I might be wrong) :

We can break up the $ \overline{A \cap B} $ into $ A \cap B$ and all L the set of all the limit points of $ A \cap B$.

The second set exist out of the closure over A and the closure over B. Those we again can split up into A and $L_A$ the set of limit points of A. The same can be done for B into B and $L_B$.

Thus we can replace $\overline{A\cap B} \subseteq \overline{A} \cap \overline{B}$ by $ (A\cap B)\cup L \subseteq (A\cup L_A)\cap(B\cup L_B)$ We can work out the left part of the equation : $(A\cap B)\cup L \subseteq (A\cap B)\cup(L_A \cap L_B)\cup(L_A \cap B)\cup (L_B \cap A)$ and then we can eliminate $(A\cap B)$ from both sides : $ L \subseteq (L_A \cap L_B)\cup(L_A \cap B)\cup (L_B \cap A)$

Is my proof correct up to here? How can I proof the remaining statement? It belonged to the section neighborhoods but I'm not quiet sure where the connection lies.

Thanks in advance.

Answer 1

If you take $x \in \overline{A \cap B}$, we have that for every open neighborhood U of $x$, $U \cap (A \cap B) \ne \emptyset. $ The proof follow from that.