# Abstract Nonsense

## Just For Fun(Rudin’s Topology Section) Part IV

28.

Problem: Prove that ever closed set in a separable metric space $\mathcal{M}$  is the union of a (possibly empty) perfect set and a set which is a set which is countable.

Proof: It is easy to see that the above proof (proof 27.) holds in any space which is second countable, in particular for separable metric spaces. Thus, if $E\subseteq\mathcal{M}$ is closed we surely have that $E\subseteq \mathfrak{C}\cup\left(\left(\mathbb{R}-\mathfrak{C}\right)\cap E\right)$. Thus, $E=\left(E\cap\mathfrak{C}\right)\cup\left(\left(\mathbb{R}-\mathfrak{C}\right)\cap E\right)$ but since $\mathfrak{C}\subseteq D(E)$ and $D(E)\subseteq E$ we have then that $E=\mathfrak{C}\cup\left(\left(\mathbb{R}-\mathfrak{C}\right)\cap E\right)$ which is the union of a perfect and countable set respectively. $\blacksquare$

29.

Problem: Prove that every open set $U\subseteq\mathbb{R}$ may be written as the disjoint union of countably many open intervals.

Proof: We need to prove a quick lemma

Lemma: Let $X$ be a topological space and let $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a class of connected subspace of $X$ such that $U_{\alpha_1}\cap U_{\alpha_2}\ne\varnothing,\text{ }\alpha_1,\alpha_2\in\mathcal{A}$ then $\displaystyle \Lambda=\bigcup_{\alpha\in\mathcal{A}}$ is a connected subspace of $X$.

Proof: Suppose that $\left(E\cap \Lambda\right)\cup\left(G\cap \Lambda\right)=\Lambda$ is a separation of $\Lambda$. We may assume WLOG that $U_{\alpha_0}\cap E\ne\varnothing$ for some $\alpha_0\in\mathcal{A}$. So, now we see that $U_{\alpha_0}\subseteq E\cap \Lambda$ otherwise $E\cap U_{\alpha_0},G\cap U_{\alpha_0}$ would be non-empty disjoint subsets of $U_{\alpha_0}$ whose union is $U_{\alpha_0}$ contradicting that $U_{\alpha_0}$ is connected. Thus, it easily follows that for any  $U_\alpha$ we have that $U_\alpha\cap U_{\alpha_0}\ne\varnothing$ so that $U_{\alpha}\cap E\ne\varnothing$ and thus by a similar reasoning we see that $U_\alpha\subseteq E\cap\Lambda$. Thus, since $\alpha$ was arbitrary it follows that $\Lambda\subseteq E\cap\Lambda$ contradicting that $G\cap\Lambda\ne\varnothing$. $\blacksquare$

So, now for each $x\in U$ define

$\mathcal{C}(x)=\left\{V\subseteq U: V\text{ is open in }U,\text{ }V\text{ is connected },\text{ and }x\in V\right\}$

And let $\displaystyle C(x)=\bigcup_{V\in\mathcal{C}(x)}V$ and finally we prove that

$\Omega=\left\{C(x):x\in U\right\}$

is a countable class of disjoint open intervals whose union is $U$. The fact that each $C(x)$ is an open interval follows since it is the union of open connected spaces with non-empty intersection (each element of $\mathcal{C}(x)$ contains $x$) it is also an open connected subspace of $\mathbb{R}$ (note that each element of $\mathcal{C}(x)$ is open in $U$ but since $U$ is open it is also open in $\mathbb{R}$. But, it was proven in the book the only connected subspace of $\mathbb{R}$ are intervals and thus $C(x)$ is an interval for each $C(x)\in\Omega$.

Now, to see that they are disjoint we show that if $C(x)\cap C(y)\ne\varnothing$ then $C(x)=C(y)$ from where the conclusion will follow. So, to see this we first note that if $C(x)\cap C(y)$ is non-empty then $C(x)\cup C(y)$ is an open connected subspace of $U$ containing both $x$ and $y$ and thus $C(x)\cup C(y)\subseteq C(x)$ and $C(x)\cup C(y)\subseteq C(y)$ but this implies that $C(y)\subseteq C(x)$ and $C(x)\subseteq C(y)$ respectively from where the conclusion follows.

Now, to see that $\Omega$ is countable we notice that for each $\Omega$ we have that $C(x)\cap\mathbb{Q}\ne\varnothing$ and so if we let $F\left(C(x)\right)$ be a fixed but arbitrary $q\in C(x)\cap\mathbb{Q}$ then

$F:\Omega\to\mathbb{Q}:C(x)\mapsto F\left(C(x)\right)$

is an injection since the elements of $\Omega$ are pairwise disjoint. The fact that $\Omega$ is countable follows immediately.

Thus, $\Omega$ is a countable collection of open intervals and

$\displaystyle U=\coprod_{C(x)\in\Omega}C(x)$

Thus, the conclusion follows. $\blacksquare$

30.

Problem: Prove that if $\displaystyle \mathbb{R}^n=\bigcup_{n\in\mathbb{N}}F_n$ where each $F_n$ is a closed subset of $\mathbb{R}^n$ then at least one $F_n$ has non-empty interior.

Proof:

Lemma: Let $\mathcal{M}$ be a complete metric space and $\left\{K_n\right\}_{n\in\mathbb{N}}$ a descending sequence of non-empty closed subsets of $\mathcal{M}$ such that $\text{diam }K_n\to 0$. Then, $\displaystyle \bigcap_{n\in\mathbb{N}}K_n$ contains one point.

Proof: Clearly $\displaystyle \bigcap_{n\in\mathbb{N}}K_n$ does not contain more than one point since $\text{diam }K_n\to 0$. So, now for each $n\in\mathbb{N}$ choose some $x_n\in K_n$ and let $P=\left\{x_n:n\in\mathbb{N}\right\}$. Now, if $P$ were somehow finite a similar argument to that used in the first case of problem 26. is applicable, so assume that $P$ is infinite. But, since $\text{diam }K_n\to 0$ it is evident that $\left\{x_n\right\}_{n\in\mathbb{N}}$ is a Cauchy sequence and thus by assumption it converges to some point $x\in\mathcal{M}$. We claim that $\displaystyle x\in\bigcap_{n\in\mathbb{N}}K_n$. To see this we note similarly to problem 26. that since $P$ is infinite it is easy to see that $x$ is a limit point of $P$ and so if $x\notin K_{n_0}$ for some $n_0\in\mathbb{N}$ then $\mathcal{M}-K_{n_0}$ is a neighborhood of $x$ containing only finitely many points of $P$ which clearly contradicts that it is a limit point. The conclusion follows. $\blacksquare$

So, now suppose each $F_n$ had empty interior (i.e. nowhere dense) . Then, since $\mathcal{M}$ is open and $F_1$ is nowhere dense there exists an open ball $B_1$ of radius less than one such that $B_1\cap F_1=\varnothing$. Let $E_1$  be the concentric closed ball of $B_1$ whose radius is half that of $B_1$. Since $F_2$ is nowhere dense $E_1^{\circ}$ contains an open ball $B_2$ of radius less than one-half which is disjoint from $F_2$.  Let $E_2$ be the concentric closed ball of $B_2$ whose radius is one-half that of $B_2$. Since $F_3$ is nowhere dense we have that $E_2^{\circ}$ contains an open ball $B_3$ of radius less than one-fourth which is disjoint from $F_3$. Let $E_3$ be the concentric closed ball of $B_3$ whose radius is half that of $E_3$. Continuing in this we get a descending sequence of non-empty closed subsets $\left\{E_n\right\}$ for which $\text{diam }E_n\to 0$. Thus, by our lemma we have that there is some $\displaystyle x\in\bigcap_{n\in\mathbb{N}}E_n$. This point is clearly not in any of the $F_n$‘s from where the conclusion follows. $\blacksquare$