Given a topological space ($X$, $\mathcal{T}$) and the following defintions
- $\partial A$ := { $x$ | every neighbourhood of $x$ contains points from $A$ and from $X \setminus A$ }
- $\overline{A}$ := $A \cup \partial A$
- $A^{\circ}$ := $A \setminus \partial A$.
Now i want to prove that $ \overline{A \cap B} \subseteq \overline{A} \cap \overline{B} $ Proof: With set theory $ \overline{A} \cap \overline{B} = (A \cup \partial A) \cap (B \cup \partial B) = (A \cap B) \cup (\partial A \cap B) \cup (A \cap \partial B) \cup (\partial A \cap \partial B) $ and $\overline{A \cap B} = (A \cap B) \cup \partial (A \cap B)$. So if $x \in \overline{A \cap B}$. Now i distinguish two cases.
(i) $x \in (A \cap B)$, then its obvious that $x \in \overline{A} \cap \overline{B}$ with the equations from above
(ii) $x \in \partial (A \cap B)$, here i have no idea how to proceed, because i have no idea how to decompose the expression $\partial (A \cap B)$, do you have any suggestions for me?