# $\partial (A \cap B) = \partial A \cap \partial B$?

let $A,B$ be subsets of topological space $X$. Does it follow

$\partial (A \cap B) = \partial A \cap \partial B$ ?

• No. There is absolutely no reason to think this would be the case. Work on your intuition. – TBrendle Oct 21 '13 at 18:44
• Try to find counterexamples for both inclusions. – Stefan Hamcke Oct 21 '13 at 18:45

Let $X = \mathbb{R}$, $A = \mathbb{Q}$, and $B = \mathbb{R} \setminus \mathbb{Q}$ be the rationals and irrationals. What do you get?
Consider $X=\mathbb{R}$, with the usual topology. If $A=[-1,1]$ and $B=[-2,2]$, then $\partial(A\cap B)=\partial([-1,1])=\{-1,1\}$; $\partial A=\{-1,1\}$; and $\partial B=\{-2,2\}$.
But certainly, $\{-1,1\}\neq \{-1,1\}\cap\{-2,2\}$.