$E$ is closed $\iff\partial E$ (boundary of set $E$) $\subseteq E$ I am studying topology of euclidean space from William Wade's text book. 
I saw this question. But I cannot come up with any ideas. 
Please show me the solution in an instructive an clear way.
Thank you for yourhelp. 
$E$ is closed $\iff\partial E$ (boundary of set $E$) $\subseteq E$

 A: Prove (if you haven't already) that a set is closed $\iff E=\overline E$.
Since $\partial E=\overline{E}\cap\overline{E^c}\subseteq \overline E=E$, the result follows. On the other hand, note (prove it) that $\overline E=E\cup \partial E$, so if $\partial E\subseteq E$, $E=\overline E$, so $E$ is closed.
ADD Given a set $E$ on a space $(X,\mathscr T)$, one can define the closure of a set to be the intersection of all closed sets that contain $E$, that is $$\overline E=\bigcap\{F\subseteq X:E\subset F\text{ and } F \text{ is closed}\}$$
That is why we usually say $\overline E$ is the smallest set (w.r.t. inclusion) that contains $E$. Because $\overline E$ is the intersection of closed sets, it is closed. Thus, if $E=\overline E$, $E$ is seen to be closed. On the other hand, if $E$ is closed, $E$ itself is a closed set containing $E$, so $\overline E\subseteq E$. Since by definition, we always have $E\subseteq\overline E$, it follows $E=\overline E$.
A: $x$ is a boundary point if every its neighborhood intersect both $E$ and $X\setminus E$.
If $E$ is closed, its complement $F=X\setminus E$ is open and every point $x\in F$ has a neighborhood contained in $F$, i.e. no points in $F$ are boundary points of $E$.
The other way around. If $\partial E\subseteq E$ then every point $x\in F$ is not boundary, i.e. it has a neighborhood that does not intersect either $E$ or $F$. But since $x\in F$, the neighborhood does not intersect $E$, and therefore, $F$ is open, $E$ is closed.
A: Hint: $\partial E=\overline E \cap \overline{X-A}$ and $\partial E⊆\overline E$. and $$ E \text{ is closed} \iff E=\overline E $$
A: $\text{Useful results}:$ Without any notion of metric spaces, the arbitrary intersection of closed sets is closed from the fact that the arbitrary reunion of open sets is open.
Indeed, for any collection of open sets $\{\mathcal O_j \}_{j\in A},\:$whenever $\:\alpha\in \mathcal O_{j^*},\:$an element of any indexed set from the arbitrary collection, we know that $\:\alpha\in\mathcal O,\:$the union, because $\mathcal O_{j^*}\:$being open means that $$\forall\alpha\in\mathcal O_{j^*}\:\exists\delta^*_{>0}\:\:\text{s.t.}\underbrace{\:(\alpha-\delta^*,\alpha+\delta^*)}_{\large \mathcal N_{\delta^*}\normalsize (\alpha)}\subset \mathcal O_{j^*}\subset \mathcal O.$$
$\implies \bigcup_{j\in A}\mathcal O_j =\mathcal O\:\:$is open $\iff \bigcap_{j\in A}\mathcal O_j^c=\mathcal O^c\:\:$is closed, by De Morgan's laws.
Another thing is that $\:(\partial E \cup \text{int}(E))\subset\overline E \implies\partial E\subset E.$
Indeed, if $\:\beta\in\left(\partial E \cup \text{int}(E)\right),\:$then either $\:\beta\in\text{int}(E)\subset E\subset \overline E\:$ or $\:\beta\in\partial E\:$ but that's it since $\:$ $(\partial E \cap \text{int}(E))=\emptyset.$
Now, $\:\beta\in \partial E\implies(\forall \delta_{>0}\:\exists\mathcal N_\delta\small(\beta)\normalsize \cap E^c\neq\emptyset\:\land\: \forall \delta_{>0}\:\exists\mathcal N_\delta\small(\beta)\normalsize \cap E\neq\emptyset),\:$ which is more restrictive than $\:\:\beta \in \overline E\implies \: (\forall \delta_{>0}\:\exists\mathcal N_\delta\small(\beta)\normalsize \cap E\neq\emptyset)$
That's why $\:\partial E\subset \overline E$
