# Abstract Nonsense

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

13.

Problem: Form a compact subset of the real numbers (with the usual topology) which has countably many limit points.

Proof: Let $\displaystyle \Omega=\left\{\frac{1}{2^n}\left(1-\frac{1}{m}\right):m,n\in\mathbb{N}\right\}$ which is tediously, but easily verified. Sorry folks, I’m not interested enough to Tex it all out. Comment me asking for a full solution if you so wish.

14.

Problem: Given an open cover $(0,1)$ (with the usual topology) which has no finite subcover.

Proof: How about $\displaystyle \left\{\left(0,1-\frac{1}{n}\right):n\in\mathbb{N}\right\}$. So, we must first show that this in fact covers $(0,1)$. So, let $x\in(0,1)$ be arbitrary and let $\delta=1-x$. Then, we have that $0<1-x$ and so by the Archimedean principle there exists some $n_0\in\mathbb{N}$ such that $\displaystyle \frac{1}{n_0}<1-x \implies x<1-\frac{1}{n_0}$ and thus $\displaystyle x\in\left(0,1-\frac{1}{n_0}\right)$. Now, to see that no finite subcover works, let $\displaystyle \left\{\left(0,1-\frac{1}{n_1}\right),\cdots,\left(0,1-\frac{1}{n_m}\right)\right\}$ be any finite subclass of of $\displaystyle \left\{\left(0,1-\frac{1}{n}\right)\right\}_{n\in\mathbb{N}}$. Then, let $N=\max\{n_1,\cdots,n_m\}$. Now, it clearly follows that $\displaystyle \left(0,\frac{1}{n_1}\right)\cup\cdots\cup\left(0,1-\frac{1}{n_m}\right)=\left(0,1-\frac{1}{N}\right)$ and so we mus merely show that there is some $\displaystyle x\in (0,1)-\left(0,1-\frac{1}{N}\right)$. But, we merely need realize that $\displaystyle 1-\frac{1}{N}<\frac{1}{2}\left(1+1-\frac{1}{N}\right)<1$ and so the conclusion follows. $\blacksquare$

15.

Problem: Show that the intersection of arbitrarily many closed subsets  of a metric space with the finite intersection property  may be empty if they are just closed, or just bounded.

Proof: Just take $\mathbb{R}$ for example. Then, the class $\left\{\mathbb{R}-B_n(0)\right\}_{n\in\mathbb{N}}$ is a class of closed subsets of $\mathbb{R}$ with the FIP but their intersection is empty. To see this we merely note that for each $x\in\mathbb{R}$ we have that $x\notin \mathbb{R}-B_{m}(x),\text{ }m=\left \lceil |x|\right\rceil +1$

Also, if we just have bounded then the claim is also wildly false in $\mathbb{R}$. Take $\left\{B_{\frac{1}{n}}(0)-\{0\}\right\}_{n\in\mathbb{N}}$. This is a class of bounded subsets of $\mathbb{R}$ with the FIP, but clearly $\displaystyle \bigcap_{n\in\mathbb{N}}\left(B_{\frac{1}{n}}(0)-\{0\}\right)=\varnothing$ since choosing $x\in\mathbb{R}-\{0\}$ we have that $\displaystyle \frac{1}{n_0}<|x|$ for some $n\in\mathbb{N}$ (by the Archimedean principle) and thus $x\notin B_{\frac{1}{n_0}}(0)-\{0\}$. $\blacksquare$

16.

Problem: Regard $\mathbb{Q}$ as a metric space with $d(p,q)=|p-q|$. Let $E=\left\{q\in\mathbb{Q}:2. Prove that a) $E$ is closed in $\mathbb{Q}$, b) it is bounded in $\mathbb{Q}$, but it is not compact in $\mathbb{Q}$.

Proof:

a) We first prove a nice little lemma (which is almost immediate from theorem 2.30)

Lemma: Let $\mathcal{M}$ be a metric space and $\mathcal{N}$ a subspace (in the metric space sense, don’t get too ahead of yourself with topological terms…we haven’t yet proven they are equivalent), then $E\subseteq\mathcal{N}$ is closed in it if and only if $E=G\cap\mathcal{N}$ for some set $G\subseteq\mathcal{M}$ which is closed in $\mathcal{M}$.

Proof: First suppose that $E\subseteq\mathcal{N}$ is open, then $\mathcal{N}-E$ is open and thus by theorem 2.30 we have that $\mathcal{N}-E=\mathcal{N}\cap\left(\mathcal{M}-G\right)$ for some set $G$ closed in $\mathcal{M}$. A little set theoretic self-convincing then shows that $E=\mathcal{N}\cap G$.

Conversely, suppose that $E=\mathcal{N}\cap G$ for some set $G$ closed in $\mathcal{M}$. Then, $\mathcal{N}-E=\mathcal{N}\cap\left(\mathcal{M}-G\right)$ (once again with a little set theoretic magic…be sure to be mindful of what set theoretic difference means here!) and thus by theorem 2.30 again we see that this means that $\mathcal{N}-E$ is open and thus $E$ closed. $\blacksquare$

From this it immediately follows that $E$ is closed in $\mathbb{Q}$ since evidently $E=\mathbb{Q}\cap[\sqrt{2},\sqrt{3}]$ and $[\sqrt{2},\sqrt{3}]$ is closed in $\mathbb{R}$.

b) It is evidently bounded.

c) We prove that $E$ is not compact in $\mathbb{Q}$ by a roundabout but instructive way.

Lemma: Let $X$ be a topological space and $Y$ a subspace. Then, $Z$ is a compact subspace of $Y$ if and only if it’s a compact subspace of $X$.

Proof: First suppose that $Z$ is a compact subspace of $Y$ and let $\left\{U_\alpha\cap Z\right\}_{\alpha\in\mathcal{A}}$ be an open cover $Z$ as a subspace of $X$ where $U_\alpha$ is open in $X$. Then, since $Y$ is a subspace of $X$ it evidently follows that $U_\alpha\cap Y$ is open in $Y$ for every $\alpha\in\mathcal{A}$. Thus, it is evident that $\left\{U_\alpha\cap Y\cap z\right\}_{\alpha\in\mathcal{A}}$ is an open cover of $Z$ as a subspace of $Y$. So, by assumption it admits a finite subcover $\left\{U_{\alpha_1}\cap Y\cap Z,\cdots,U_{\alpha_n}\cap Y\cap Z\right\}$ and so it readily follows that $\left\{U_{\alpha_1}\cap Z,\cdots,U_{\alpha_n}\cap Z\right\}$ is a finite subcover of $\left\{U_\alpha\cap Z\right\}_{\alpha\in\mathcal{A}}$

Conversely, suppose that $Z$ is compact as a subspace of $X$ and $\left\{U_{\alpha}\cap Z\right\}_{\alpha\in\mathcal{A}}$ is an open cover for $Z$ as a subspace of $Y$. Then, since $Y$ is a subspace of $X$ we have that $U_\alpha=V_\alpha\cap Y$ for some open $V_\alpha$ in $X$. Thus, it is evident that $\left\{V_\alpha\cap Z\right\}_{\alpha\in\mathcal{A}}$ is an open cover for $Z$ as a subspace of $X$ and so by assumption it admits a finite subcover $\left\{V_{\alpha_1}\cap Z,\cdots,V_{\alpha_n}\cap Z\right\}$ from where a quick check shows then that $\left\{U_{\alpha_1}\cap Z,\cdots,U_{\alpha_n}\cap Z\right\}$ is the desired finite subcover of $\left\{U_\alpha\cap Z\right\}_{\alpha\in\mathcal{A}}$. $\blacksquare$

From the above lemma we see that $E$ will be compact as a subspace of $\mathbb{Q}$ if and only if it is compact as a subspace of $\mathbb{R}$. But, it is not since evidently $\sqrt{2}$ is a limit point of $E$ but not a point of $E$, and thus $E$ is not closed and thus not compact. $\blacksquare$

17.

Problem: Let $E=\left\{x\in[0,1]:x=.a_1a_2a_3\cdots,\text{ }a_k=4,7\right\}\subseteq [0,1]$; a) Is $E$ countable? b) Is $E$ dense in $[0,1]$? c) Is $E$ dense in $[0,1]$? d) Is $E$ perfect in $[0,1]$?

Proof:

a) No, $E$ is not countable. It is evident that $E\simeq\{0,1\}^{\mathbb{N}}\simeq\mathcal{P}(\mathbb{N})\simeq\mathbb{R}$. The only one that may be slightly hazy is that $\{0,1\}^{\mathbb{N}}\simeq\mathcal{P}(\mathbb{N})$ but it is pretty routine to check that $\eta:\{0,1\}^\mathbb{N}\to\mathcal{P}(\mathbb{N}):f\mapsto f^{-1}(\{1\})$ is a bijection. (this is just the correspondence between subsets and indicator functions).

b) Of course it isn’t dense. It is clear that $\left(0,\tfrac{1}{3}\right)\cap E=\varnothing$.

c) It suffices from prior comment to show that it’s compact as a subspace, and by virtue of the Heine-Borel theorem we must merely show that $E$ is closed and bounded. Boundedness is clear though since $\displaystyle E\subseteq\left[\frac{4}{9},\frac{7}{9}\right]$. So, suppose that $x\in D(E)$ but $x\notin E$. Then, we have that $x=.b_1b_2\cdots$ where $b_k\ne4,7$ for some $k\in\mathbb{N}$. But, note then that for any $e\in E$ we have that $|e-x|\geqslant\frac{1}{10^k}$ and thus $B_{\frac{1}{10^k}}(x)\cap E=\varnothing$ which of course is a contradiction. Thus, the conclusion follows.

d) It suffices to prove (since we proved in c) that $E$ is closed) that $E\subseteq D(E)$. To do this let $x\in E$ and let $\delta>0$ be given. Then, by the Archimedean principle there exists some $n\in\mathbb{N}$ such that $\displaystyle \frac{3}{10^n}<\delta$ and so if $x=.a_1a_2\cdots$ then we have that

$e=.b_1b_2\cdots,\text{ }b_k=\begin{cases} a_k \quad\text{if}\quad k\ne n \\ 4,7\quad\text{if}\quad k=n\end{cases}$

Where obviously we choose $4,7$ to be the opposite of what $a_n$ is. Then, $\displaystyle |e-x|=\frac{3}{10^n}<\delta$ and $e\ne x$. It follows that $x\in D(E)$.

18.

Problem: Is there a non-empty perfect subset of $\mathbb{R}$ (with the usual topology) which contains no rational values?

Proof: The answer is yes, but to say it right takes some measure theory. We know that $\mu(\mathbb{Q})=0$ (the usual measure) and so there exists countably many open intervals $\mathcal{I}_n$ such that $\displaystyle \mathbb{Q}\subseteq\bigcup_{n\in\mathbb{N}}\mathcal{I}_n$ and $\displaystyle \sum_{n=1}^{\infty}\mu\left(\mathcal{I}_n\right)<1$. So, we have that $\displaystyle \mathbb{R}-\bigcup_{n\in\mathbb{N}}\mu\left(\mathcal{I}_n\right)$ is closed, non-empty (since it is non-zero measure). So, once again since it has positive measure it must be uncountable and by problem 28 it follows that the set of all condensation points of $\displaystyle \mathbb{R}-\bigcup_{n\in\mathbb{N}}\mathcal{I}_n$ is perfect, and since each condensation point is a limit point it must be a point of $\displaystyle \mathbb{R}-\bigcup_{n\in\mathbb{N}}\mathcal{I}_n$ and thus not a rational number. $\blacksquare$

19.

Problem: a) If $A,B$ are disjoint closed sets in a metric space $\mathcal{M}$ then they are separated. b) If $A,B\subseteq\mathcal{M}$ are disjoint and open they are separated. c) Fix $x_0\in\mathcal{M},\delta>0$ and define $A=\left\{y\in \mathcal{M}:d(x_0,y)<\delta\right\}$ and $B=\left\{y\in\mathcal{M}:d(x_0,y)>\delta\right\}$, and prove that $A,B$ are separated. d) Prove that every connected metric space with more than one point  is uncountable.

Proof:

a) Clearly $A,B$ are separated since $\overline{A}\cap B=A\cap\overline{B}=A\cap B=\varnothing$.

b) Suppose that $A\cap\overline{B}\ne\varnothing$ then it must follow that there is some $x\in A\cap D(B)$. But, since $x\in D(B)$ every neighborhood $N$ of $x$ intersects $B$ and thus intersects $\mathcal{M}-A$ and thus $x\notin A^{\circ}=A$ which is a contradiction. An identical argument works for the other cases.

c) Merely note that $\varphi:\mathcal{M}\to \mathbb{R}:y\mapsto d(x_0,y)$ is continuous (since it’s the normal metric restricted to $\{x_0\}\times \mathcal{M}\approx \mathcal{M}$). And, that $A=\varphi^{-1}((-\infty,\delta)),B=\varphi^{-1}((\delta,\infty))$ which are disjoint and open. The conclusion follows from b)

d) Suppose that $\mathcal{M}$ were countable and has more than one point, then so is $\mathcal{M}\times \mathcal{M}$ and then so is $\displaystyle d\left(\mathcal{M}\times \mathcal{M}\right)=\left\{d(x,y):x,y\in \mathcal{M}\right\}$. So, choose $x_0,y_0\in \mathcal{M}$ to be distinct and let $\xi=d(x_0,y_0)>0$. But, since $(0,\xi)\subseteq\mathbb{R}$ is uncountable and $d\left(\mathcal{M}\times\mathcal{M}\right)$ countable there exists some $\delta\in (0,\xi)-d\left(\mathcal{M}\times\mathcal{M}\right)$. So, let $A=\left\{y\in\mathcal{M}:d(x_0,y)<\delta\right\}$ and $B=\left\{y\in\mathcal{M}:d(x_0,y)>\delta\right\}$. Then, $A,B$ are non-empty ($0\in A$ and $y_0\in B)$ disjoint and since $d(x_0,y)\ne\delta,\text{ }y\in\mathcal{M}$ their union is equal to $\mathcal{M}$. It follows that $\mathcal{M}$ is not connected. $\blacksquare$

20.

Problem: Are closures and interiors of connected sets always connected?

Proof: Closures are, we reprove this fact for general topological spaces.

Lemma: Let $X$ be a topological space, $E$ a connected subspace and $E\subseteq G\subseteq\overline{E}$ then $G$ is a connected subspace.

Proof: Evidently $\text{cl}_G\text{ }E=G$. So, let $\varphi:G\to D$ (where $D$ is the two-point discrete space) be continuous. Then since $E$ is connected must have that $\varphi\mid_{E}$ is constant, but since $E$ is dense in $G$, $\varphi$ continuous and $D$ Hausdorff it follows that $\varphi$ is constant. The conclusion follows. $\blacksquare$

The interior of connected subspaces is not always connected though. Consider the “dumbbell” shape $B_{1}[-2]\cup[-1,1]\times\{0\}\cup B_{1}[2]$. This is clearly connected (it is the union of three connected sets with non-empty intersection) but it’s interior is $B_{1}(-2)\cup B_{1}(2)$ which is as classically disconnected as one can get. $\blacksquare$

Remark: It is interesting to note that the result is true for $\mathbb{R}$, since the only connected subsets are interiors which (as can easily be checked by case) the interiors of which are intervals.