In topology, any relationship on boundary like $Bd(A\cap B)$ and $Bd(A) \cap Bd(B)$? Given a topological space $(X,\mu)$, let $Bd(S)$ denote the boundary of subset $S\subseteq X$.
Is there any relationship between $Bd(A\cap B)$ and $Bd(A)\cap Bd(B)$ for arbitrary subsets $A,B\subseteq X$?
This question came to my mind when I did the following assignment from topology lecture:

Prove $Bd(A\cup B)\subset Bd(A)\cup Bd(B)$.

After completing the proof I thought "how about the intersection?"
It would be great to consider the above in infinite sets and finite sets, if the results are different.
 A: There's no containment relationship between $\partial A\cap\partial B$ and $\partial(A\cap B)$ that always holds.  To prove that we need two counter-examples.
If $A=\mathbb Q$ and $B=\mathbb R\setminus\mathbb Q$, then $\partial A=\partial B=\mathbb R$.  But $\partial (A\cap B)=\emptyset$.  So it's not true in general that $\partial A\cap\partial B\subset\partial(A\cap B)$.
On the other hand if $A=\mathbb Q$ and $B=\mathbb R$, then $\partial A=\mathbb R$.  And $A\cap B=A$.  So $\partial (A\cap B)=\mathbb R$.  But $\partial B=\emptyset$.  So $\partial A\cap\partial B=\emptyset$.  Thus it's not true in general that $\partial A\cap\partial B\supset\partial(A\cap B)$.
So there is no general statement you can make.  But maybe you can put some restrictions on $A$ and $B$ so that something is true in some special case.
A: If we take $Bd(A) = \overline{A}\backslash A^{\circ}$, then $$Bd(\mathbb Q) = \mathbb R,$$ and $$Bd(\mathbb R \backslash \mathbb Q) = \mathbb R,$$ so that $$Bd(\mathbb Q)\cap Bd(\mathbb R \backslash \mathbb Q) = \mathbb R.$$
But $Bd(\mathbb R \backslash \mathbb Q \cap \mathbb Q) = Bd(\emptyset) = \emptyset$.
So $Bd(\mathbb R \backslash \mathbb Q \cap \mathbb Q) \subset Bd(\mathbb Q)\cap Bd(\mathbb R \backslash \mathbb Q)$.
Now use $A = (0,1)$ and $B=(1/2,2)$ to see that this relation fails in general. 
In fact, $Bd(A\cap B) = \{1/2,1\}$ whereas $Bd(A)\cap Bd(B) = \emptyset$.
So no simple relation exists in general.
