Let $E$ be a subset of a metric space $(S,d)$.
Prove that:
A point is in the boundary of $E$ if and only if it belongs to the closure of both $E$ and its complement.
Here is what I thought:
I'm first trying to understand what I need to prove.
The boundary of $E$ is the set $E^- -E ^{\circ}$.
It a point belongs to both $E^-$ and $(S-E)^-$, then it belongs to $E^- \cap (S-E)^-$.
Therefore I think I need to prove that:
$$E^- -E ^{\circ}=E^- \cap (S-E)^-$$
And now I'm not sure if there any set theory rules I could use.
Intuitive I would say that I need to prove: $S-E ^\circ =(S-E)^-$. Is this correct ?
And I can kind of feel that this last statement is right, but I can't prove this rigoursly.
Could anybody help me how I can prove this statement more rigourisly ?
Best Answer
You're absolutely right!
Note that if a point $x$ is interior to $E$, then it certainly isn't in $S-E$. In fact there is some $r>0$ such that any point of $S-E$ is separated from $x$ by a distance of at least $r$, so $x$ isn't a limit point of $S-E$, either. Hence, $E^o$ and $(S-E)^-$ are disjoint, so that $S-E^o$ is contained in $(S-E)^-$.
On the other hand, suppose $x\in(S-E)^-$. If $x\in S-E$, then $x\notin E$, so $x\notin E^o\subset E,$ and so $x\in S-E^o$. Otherwise, $x$ is a limit point of $S-E$, so for all $r>0$ there is a point $y_r\in S-E$ such that $0<d(x,y_r)<r$. Hence, no open ball around $x$ is disjoint from $S-E$, so no open ball around $x$ is contained in $E$, so $x\notin E^o$, and so $x\in S-E^o$. Thus, we have the other inclusion.