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I'm having trouble verifying my proof, would appreciate some input on this one.

Let $(X,d)$ be a metric space with $E\subset X$.

Suppose $E$ is closed in $X$, which means that $E=\overline{E}$. By defintion of the boundary of $E$, denoted $\partial E$ we have that

$\partial E=\overline{E}\cap\overline{E^{c}}\subseteq \overline{E}=E$ which easily establishes the first part.

Moving on to the converse, where I'm having my main doubts, we assume that

$\partial E \subseteq E$ or equivalently $\overline{E}\cap \overline{E^{c}}\subseteq E$

Now $\forall x\in E^{c}$ we have that $x \notin \partial E$.

Noticing that $\partial E$ is closed, since it is an intesection of two closed sets.

I can't get any further than this... would appreciate a detailed help of completing the last part. Thanks

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up vote 0 down vote accepted

Note that $x \in \partial E$ iff every open $U$ containing $x$ intersects both $E$ and $E^c$.

Suppose $x \notin E$, and so $x \notin \partial E$. Hence there is some open $U$ containing $x$ that does not intersect one of $E$ or $E^c$. Since $x \in E^c$, we must have $U \cap E = \emptyset$. Hence $E^c$ is open.

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Hint: Assume there is an $x\in E^c$ that fulfills $x\in\delta E$ and show that then $E$ is not closed

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Suppose there exists $x \in \partial E$ with $x \in E^c$. Every neighborhood of $x$ contains points of $E$. Thus $x$ is not interior to $E^c$.

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Hope it will give you an easier proof.

Try to visualize it geometrically. We know a closed $E$ set contains all its limit point. For any point $a$ on the boundary we get a sequence $\{x_n\}$ having infinitely many distinct point in $E$ converging to $a$. So $a$ is a limit point of $E$. Thus the boundary is contained in $E$.

Consider any limit point of $E$ say $e$. If $e \in E$ $E$ is trivially closed. If $e$ does not belong to $E$, you shall get a nbd of $e$ which contains both the points of $E$ and $E'$. See $e$ is on the boundary of $E$. Now if you consider that the boundary is contained in $E$, you can immediately show that $E$ is closed.

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