# $E$ 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 $E$) $\subseteq E$

• How do you define $E$ to be closed? If you define the closure of $E$ to be $\overline E=E\cup\partial E$, and if you have either proven or defined that a set is closed if $\overline {E}=E$; then the conclusion is immediate.
– Pedro
Commented Jun 19, 2013 at 17:35
• Sorry I am confused. Now you show the first part of the proof. Is ıt right? @PeterTamaroff
– 1190
Commented Jun 19, 2013 at 17:40
• Sorry, I know I dont want such thing. But, please can you show me this more teachable below answer part? I want to learn. Please:) @PeterTamaroff
– 1190
Commented Jun 19, 2013 at 17:42
• @B11b, the question is what definition of "closed" did you learn? There are several equivalent definitions. Commented Jun 19, 2013 at 17:43
• @B11b As Yoni says, we need to know how you define a set to be closed, or what theorems you can use. Depending on that, the proof can be more or less contorted.
– Pedro
Commented Jun 19, 2013 at 17:45

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 $E=\overline E$.

• Now, I have comprehended all thing. I tired to realized in my mind. But something was not clarified. This addition helps:) Thank you for addition:)
– 1190
Commented Jun 19, 2013 at 17:54
• Sorry, I stack with a point. $\parital E =$$E^{\overline} \E^o =E^{\overline}\cap(E^{o})^{c} Is this writing is false? – 1190 Commented Jun 19, 2013 at 22:15 • @B11b Could you rewrite that? You want \overline{E} – Pedro Commented Jun 19, 2013 at 22:17 • Sorry I write a few minitue – 1190 Commented Jun 19, 2013 at 22:19 • So sorry. I dont know to write latex enough and fast. So I added a picture above :) – 1190 Commented Jun 19, 2013 at 22:24 x is a boundary point if every its neighborhood intersect both E and X\setminus E. If E is closed, its complement F=X\setminus E is open and every point x\in F has a neighborhood contained in F, i.e. no points in F are boundary points of E. The other way around. If \partial E\subseteq E then every point x\in F is not boundary, i.e. it has a neighborhood that does not intersect either E or F. But since x\in F, the neighborhood does not intersect E, and therefore, F is open, E is closed. Hint: \partial E=\overline E \cap \overline{X-A} and \partial E⊆\overline E. and$$ E \text{ is closed} \iff E=\overline E $$• \bar seems to work fine on small stuff, but \overline is better for bigger entries-. – Pedro Commented Jun 19, 2013 at 17:42 \text{Useful results}: Without any notion of metric spaces, the arbitrary intersection of closed sets is closed from the fact that the arbitrary reunion of open sets is open. Indeed, for any collection of open sets \{\mathcal O_j \}_{j\in A},\:whenever \:\alpha\in \mathcal O_{j^*},\:an element of any indexed set from the arbitrary collection, we know that \:\alpha\in\mathcal O,\:the union, because \mathcal O_{j^*}\:being open means that$$\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\$