Exercise 1.2.3 (Basic properties of open and closed sets)

Question

Exercise 12.2.3: Prove Proposition 12.2.15.\par 

	extbf{Proposition 12.2.15}. Let $(X, d)$ be a metric space.
\begin{enumerate}[label=(\alph*)]
\item Let $E$ be a subset of $X$. Then $E$ is open if and only if $E=\operatorname{Int}(E)$. In other words, $E$ is open if and only if for every $x \in E$, there exists an $r>0$ such that $B(x, r) \subseteq E$.

\item Let $E$ be a subset of $X$. Then $E$ is closed if and only if $E$ contains all of its adherent points. In other words, $E$ is closed if and only if for every convergent sequence $\left(x_{n}ight)_{n=m}^{\infty}$ in $E$, the limit $\lim _{n ightarrow \infty} x_{n}$ of that sequence also lies in $E$.

\item For any $x_{0} \in X$ and $r>0$, the ball $B\left(x_{0}, right)$ is an open set. The set $C=\{x \in X$ : $\left.d\left(x, x_{0}ight) \leq right\}$ is a closed set.

\item Any singleton set $\left\{x_{0}ight\}$, where $x_{0} \in X$, is automatically closed.

\item If $E$ is a subset of $X$, then $E$ is open if and only if the complement $X \backslash E$ is closed.

\item If $E_{1}, E_{2}, \ldots, E_{n}$ is a finite collection of opens sets in $X$, then $E_{1} \cap E_{2} \cap \cdots \cap E_{n}$ is also open. If $F_{1}, F_{2}, \ldots, F_{n}$ is a finite collection of closed sets in $X$, then $F_{1} \cup F_{2} \cup \cdots \cup F_{n}$ is also closed.

\item If $\left\{E_{\alpha}ight\}_{\alpha \in I}$ is a collection of open sets in $X$ (where the index set $I$ could be finite, countable, or uncountable), then the union $\bigcup_{\alpha \in I} E_{\alpha}$ is also open. If $\left\{F_{\alpha}ight\}_{\alpha \in I}$ is a collection of closed sets in $X$, then the intersection $\bigcap_{\alpha \in I} F_{\alpha}$ is also closed.

\item If $E$ is any subset of $X$, then $\operatorname{Int}(E)$ is the largest open set which is contained in $E$; in other words, $\operatorname{Int}(E)$ is open, and given any other open set $X \subseteq E$, we have $V \subseteq \operatorname{Int}(E)$. Similarly $E$ is the smallest closed set which contains $E$; in other words, $\bar{E}$ is closed, and given any other closed set $K \supset E, K \supset E$.
\end{enumerate}

Prakash Anand · Accepted Answer

\begin{proof} (a) First note that if $E=\emptyset$ then we are done. Since then we would have $E=\emptyset \subseteq$ Int $(E)$. And from Lemma 1(b) we would then $E=\operatorname{Int}(E)$. So we can assume $E$ is not empty.\par $(\Rightarrow)$ Suppose $E$ is open. Then $\partial E \cap E=\emptyset$.\par Let $x \in E$, so $x otin \partial E$. Assume that $x \in \operatorname{ext}(E)$. Then $\exists r>0$ such that $B(x, r) \cap E=\emptyset$. But $x \in E$ and $x \in B(x, r) \Rightarrow B(x, r) \cap E eq \emptyset$ a contradiction. Thus, $x otin \operatorname{Ext}(E)$, hence $x \in \operatorname{Int}(E)$. That is, $E \subseteq \operatorname{Int}(E)$. And from Lemma we have $\operatorname{Int}(E) \subseteq E$. Thus $E=\operatorname{Int}(E)$.\par $(\Leftarrow)$ Suppose $E=\operatorname{Int}(E)$. Assume $\partial E \cap E eq \emptyset$ then $\exists y \in \partial E \cap E \Rightarrow y \in \partial E, y \in E \Rightarrow y \in$ Int $(E)$ a contradiction, since $\partial E \cap \operatorname{Int}(E)=\emptyset$. Thus $\partial E \cap E=\emptyset$. Therefore $E$ is open.\par (b) $(\Rightarrow)$ Suppose $E$ is closed. Then $\partial E \subseteq E$, and from Lemma $1(\mathrm{~b}) \operatorname{Int}(E) \subseteq E$, thus $\partial E \cup$ $\operatorname{Int}(E) \subseteq E .$ $(\Leftarrow)$. Suppose $\partial E \cup \operatorname{Int}(E) \subseteq E$ but then $\partial E \subseteq E$, hence $E$ is closed. \par (c) First we will prove that for any $x_{0}$ and $r>0$ that $B\left(x_{0}, r ight)$ is open. We will do this by showing that $B\left(x_{0}, r ight)=\operatorname{Int}\left(B\left(x_{0}, r ight) ight)$.\par From Lemma 1 (b) we know that $\operatorname{Int}\left(B\left(x_{0}, r ight) ight) \subseteq B\left(x_{0}, r ight)$.\par Let $x \in B\left(x_{0}, r ight)$ so $d\left(x, x_{0} ight)0$. Let $y \in B\left(x, r^{\prime} ight)$ then $d(y, x)0 \exists N>0$ such that $d\left(x_{n}, x^{\prime} ight)<\epsilon \forall n \geq N$.\par From the triangle inequality we have $d\left(x^{\prime}, x_{0} ight) \leq d\left(x^{\prime}, x_{n} ight)+d\left(x_{n}, x_{0} ight) \leq \epsilon+r$. Letting $\epsilon$ go to zero we then get that $d\left(x^{\prime}, x_{0} ight) \leq r$ hence $x^{\prime} \in C$. Therefore $C$ is closed.\par (d) There is only one convergent sequence in $\left\{x_{0} ight\}$ and that is $\left(x_{0}, x_{0}, \ldots ight)$ and this sequence converges to $x_{0} \in\left\{x_{0} ight\}$. Thus $\left\{x_{0} ight\}$ is closed.\par (e) $(\Rightarrow)$ Suppose $E$ is open, then $E=\operatorname{Int}(E)$. But, $$ \begin{aligned} &X=\operatorname{Int}(E) \cup \operatorname{Ext}(E) \cup \partial E \ &\Rightarrow X \backslash E=\operatorname{Ext}(E) \cup \partial E \ &\Rightarrow X \backslash E=\operatorname{Ext}(E) \cup \partial(X \backslash E) \ &\Rightarrow \partial(X \backslash E) \subseteq X \backslash E \end{aligned} $$ (From Lemma 1(c))\par (Since $E=\operatorname{Int}(E)$ )\par (From Lemma $1(\mathrm{~d})$ )\par Hence $X \backslash E$ is closed. $(\Leftarrow)$ Suppose $X \backslash E$ is closed. Then we have $\partial(X \backslash E) \subseteq X \backslash E .$ But from Lemma $1(\mathrm{~d})$ $\partial(X \backslash E)=\partial(E)$. Hence $\partial(E) \subseteq X \backslash E$. So if $x \in \partial E$ then $x \in X \backslash E \Rightarrow x \in X$ and $x otin E$. Hence $\partial E \cap E=\emptyset$. Therefore $E$ is open.\par (f) (i) Let $x \in E_{1} \cap \cdots \cap E_{n}$. Thus $x \in E_{i} \forall i=1, \ldots, n$. But $E_{i}$ is open for all $i$, so $\exists r_{i}>0$ such that $B\left(x, r_{i} ight) \subseteq E_{i}$. Let $r^{\prime}=\min \left\{r_{1}, r_{2}, \ldots, r_{n} ight\}$. Then $B\left(x, r^{\prime} ight) \subseteq E_{i} \forall i=1, \ldots, n$. Hence $B\left(x, r^{\prime} ight) \subseteq E_{1} \cap \cdots \cap E_{n}$. Hence $E_{1} \cap \cdots \cap E_{n}$ is open.\par (ii) From part (e) we know that $X \backslash F_{1}, \ldots, X \backslash F_{n}$ are each open. And so from part (i) of this exercise we have that $\left(X \backslash F_{1} ight) \cap \cdots \cap\left(X \backslash F_{n} ight)$ is also open. But $\left(X \backslash F_{1} ight) \cap \cdots \cap\left(X \backslash F_{n} ight)=$ $X \backslash\left(F_{1} \cup \cdots \cup F_{n} ight)$ from De Morgan's Law, and hence is also open. Thus we get that $F_{1} \cup \cdots \cup F_{n}$ is closed (from part (e) again). (g) (i) Let $x \in \bigcup_{\alpha \in I} E_{\alpha}$. That is, $x \in E_{\alpha}$ for some $\alpha \in I$. But $E_{\alpha}$ is open $\forall \alpha \in I$, hence $\exists r>0$ such that $B(x, r) \subseteq E_{\alpha} \subseteq \bigcup_{\alpha \in I} E_{\alpha}$. Thus, $\bigcup_{\alpha \in I} E_{\alpha}$ is open.\par (ii). Let $\left(x_{n} ight)$ be a sequence of elements in $\bigcap_{\alpha \in I} F_{\alpha}$ that converges to $x \in X$. Since $\forall j x_{j} \in$ $F_{\alpha} \forall \alpha \in I$ and that $F_{\alpha}$ is closed for all $\alpha \in I$ we know that $\left(x_{n} ight)$ converges in $F_{\alpha} \forall \alpha \in I$. That is, $x \in F_{\alpha} \forall \alpha \in I$. Hence $x \in \bigcap_{\alpha \in I} F_{\alpha}$. Thus, $\bigcap_{\alpha \in I} F_{\alpha}$ is closed.\par Note that you can also prove this by using DeMorgan's laws and part (i).\par (h) (i) First we will show that for $V \subseteq E, V$ open, that $V \subseteq \operatorname{Int}(E)$. Suppose $V$ is open, then $V=\operatorname{Int}(V)$. Let $x \in V=\operatorname{Int}(V)$, then $\exists r^{\prime}>0$ such that $B\left(x, r^{\prime} ight) \subseteq V \subseteq E$. But this means that $x \in \operatorname{Int}(E)$. Hence $V \subseteq \operatorname{Int}(E)$.\par Now we will show that $\operatorname{Int}(E)$ is open. Let $x \in \operatorname{Int}(E)$. Then $\exists r>0$ such that $B(x, r) \subseteq E$. But from part (c) we know that $B(x, r)$ is open $\forall x, r>0$. But that means that $B(x, r) \subseteq$ $\operatorname{Int}(E)$. Thus $\operatorname{Int}(E)$ is open.\par (ii) From Lemma 1 (e) we know that $\operatorname{Int}(X \backslash E)=X \backslash \bar{E}$. From part (i) we know that Int $(X \backslash E)$ is open, and now if we take the complement, we can use the result from part (e) and we get that $\bar{E}$ is closed.\par Let $V \supseteq E, V$ be a closed set. Hence $X \backslash V$ is open, and $X \backslash V \subset X \backslash E$. Now using part (i) of this exercise we know that $X \backslash V \subseteq \operatorname{Int}(X \backslash E)$. But from Lemma 1 (e) we know $\operatorname{Int}(X \backslash E)=X \backslash E$. Hence we get $X \backslash V \subseteq X \backslash \bar{E}$. Therefore, $\bar{E} \subseteq V$.\par \end{proof}

Exercise 1.2.3 (Basic properties of open and closed sets)

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Exercise 1.2.3 (Basic properties of open and closed sets)

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