boundary of multiple sets Is it possible for a boundary point of a set to be a boundary point with respect to multiple other sets, in the sense that any neighborhood of a point of A contains points of BOTH B and C? If so can you give me an example? If so is it then the case that an entire boundary set can be a boundary of A with both B and C? If so what are the rules governing closed and open sets? If so, is there any necessary relation implied between B and C? 
 A: A fairly famous example is the Lakes of Wada.
The Lakes of Wada are 3 pairwise disjoint open subsets of $\mathbb{R}^2$ with the property that they all have the same boundary.  Further, according the wikipedia article, a similar construction yields countably many pairwise disjoint open sets all having the same boundary.
A: So you want $\partial(A)\subseteq \partial(B)\cap\partial(C)$? 
Yes, it's possible, even with (though this was not specified) $A$, $B$, and $C$ pairwise distinct.
For example, take $X$ to be the real line with the usual topology, $A=[0,1]$, $B=(-1,0)\cup (1,2)$, $C=(0,1)$. Then $\partial(A)=\partial(C)\subseteq \partial(C)$.
I'm not sure what "what are the rules governing closed and open sets" means: the usual rules, I would guess: a set is open if and only if its complement is closed, open sets must include $\varnothing$, the entire set, and be closed under pairwise intersections and under arbitrary unions.
I don't think there is any relationship implied between $B$ and $C$. Above you see an example where $B\cap C=\varnothing$. Replace $C$ with $(-\frac{1}{2},0)\cup (\frac{1}{2},1)$ for an example where $B\cap C\neq\varnothing$, but you have neither $B\subseteq C$ nor $C\subseteq B$. And of course, if $B$ and $C$ work, then so do $X-B$ and $C$, $B$ and $X-C$, and $X-B$ and $X-C$ (since the boundary of a set equals the boundary of its complement). 
A: In the real line, you might like to consider the sets $\mathbb{Q}$, together with $\{a+\sqrt{2} : a\in\mathbb{Q}\}$ and $\{a+\sqrt{3} : a\in\mathbb{Q}\}$.
They are disjoin and each one has as boundary $\mathbb{R}$. Although they are not open or closed, they do seem to satisfy the conditions in the first part of your question.
