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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 ?

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Do you mean $E^-$ to be $\bar{E}$ = \bar{E}? or $\overline{E}$ = \overline{E} – amWhy Apr 3 '13 at 15:49
This really comes down to element chasing. Let $x$ be in the closer of $E$ but not in the interior of $E$, then it's certainly in the closure of $E$. Can you show that it's also in the closure of the complement of $E$? (Hint: it's not in the interior of $E$ so all neighbourhoods of $x$ intersect the complement of $E$). This gives you one inclusion. Try the other inclusion using similar element chasing methods. – Dan Rust Apr 3 '13 at 15:51
@amWhy In my book: Elementary Analysis by Ross, they denote the closure of the set $E$ with $E^-$. What is the standard notation for this ? – Kasper Apr 3 '13 at 15:55
Kasper, that's fine, I was just checking. Often closure of a set $E$ is denoted $\overline{E}$. – amWhy Apr 3 '13 at 16:11
up vote 4 down vote accepted

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.

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thanks ! ${}{}{}{}{}{}$ – Kasper Apr 3 '13 at 16:08

Yes. $S-E^\circ=\overline{S-E}$ is correct.

$x\notin E^\circ \iff $ there is no whole ball around $x$ within $E$ $\iff$ all balls around $x$ intersect $S-E$ $\ \iff x\in\overline{S-E}$.

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