Partitioning a metric space with interior, exterior and boundary of a set Given a set $A$ in a metric space $(X,d)$, why do its interior $A^{\circ}$, its complement's interior ($X$$\setminus$$A)^{\circ}$ and its boundary $\partial$$A$ form a partition of the whole space? I saw this was presented as a given fact in a few posts but no one really proved it. 
My try: First, I proved the three are disjoint. Then, I defined an open ball $B(x, \epsilon)$ around an arbitrary point $x$ $\in$ $X$. If there exists $\epsilon$ $\gt$ $0$ for which the ball is contained in $A^{\circ}$, then $x$ $\in$ $A^{\circ}$. Same for $x$ $\in$ ($X$$\setminus$$A)^{\circ}$ (if there exists $\epsilon$ $\gt$ $0$ such that the ball is contained in ($X$$\setminus$$A)^{\circ}$ then $x$ $\in$ ($X$$\setminus$$A)^{\circ}$). I just can't seem to find a way to prove that: if there isn't such an $\epsilon$, $x$ has to satisfy $x$ $\in$ $\partial$$A$.
Any help would be appreciated.
 A: I made an error in my comment. The correct argument is without the interior.
Let $x\in X$ such that $x\notin A^\circ$ and $x\notin (X\setminus A)^\circ$. Then, for every $\epsilon > 0$ we have
$$ B(x, \epsilon) \not\subseteq A $$
and 
$$ B(x, \epsilon) \not\subseteq X\setminus A. $$
That is, there exists some 
$$ y\in B(x,\epsilon) \cap (X\setminus A) $$ 
and 
$$ z\in B(x,\epsilon) \cap (X\setminus (X \setminus A)) = B(x,\epsilon) \cap A.$$ 
That ist, $x\in \partial A$.  
Note: Obviously, $B(x, \epsilon)$ can be replaced by any neighborhood of $x$. Thus, the proof and the statement is valid in any topological space.
A: Actually it's not just for metric spaces, but for arbitrary topologies.
Given any point $x\in X$, there are exactly three possibilites:


*

*Either there exists a neighbourhood of $x$ that contains only points from $A$. Then the point is in $A^\circ$.

*Or there exists a neighbourhood of $x$ that contains only points from $X\setminus A$. Then the point is in $(X\setminus A)^\circ$.

*Or there exists neither. Then the point is in $\partial A$.
It is obvious that at least one of the options always holds, and also that the last option cannot hold at the same time as one of the first two. So all that remains to show is that the first two cannot hold at the same time.
But that is also easy: Either $x\in A$ or $x\in X\setminus A$. Since every neighbourhood of $x$ contains $x$, each of those cases excludes exactly one of the first two options.
