# Abstract Nonsense

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

21.

Problem: Let $A$ and $B$ be separated subsets of $\mathbb{R}^n$, suppose that $\bold{a}\in A,\bold{b}\in B$ and define

$\bold{p}:\mathbb{R}\to\mathbb{R}^n:t\mapsto (1-t)\bold{a}+t\bold{b}$

Let $A_0=\bold{p}^{-1}(A),B_0=\bold{p}^{-1}(B)$. a) Prove that $A_0,B_0$ are separated subsets of $\mathbb{R}$. b) Prove that there exists $t_0\in(0,1)$ such that $\bold{p}(t_0)\notin A\cup B$. c) Prove that every convex subset of $\mathbb{R}^n$ is connected.

Proof:

a) Clearly we have that $A_0\cap B_0=\varnothing$ and so, if we assume that $x\in\overline{A_0}\cap B_0$ then we must have that $x\in D(A_0)\cap B_0$. So, we may choose $\{x_n:n\in\mathbb{N}\}\subseteq A_0$ such that $x_n\to x$ and so it easily follows that $\bold{p}(x_n)\to\bold{p}(x)$ and thus $\bold{p}(x)\in\overline{A}$ so $\bold{p}(x)\notin B$ and thud $x\notin B_0$ which is a contradiction.

b) We must merely note that if $\bold{p}((0,1))\subseteq A\cup B$ then it would be a path from $\bold{a}\in A$ to $\bold{b}\in B$ which is impossible since they are separated.

c) We can prove this more generally. Let $\mathcal{V}$ be a normed vector space and let $C\subseteq\mathcal{V}$ be convex, then $C$ is path connected. This clearly follows since the straight line $L(\bold{x},\bold{y})=\left\{t\bold{x}+(1-t)\bold{y}:t\in[0,1]\right\}$ is a path from $\bold{x}$ to $\bold{y}$.

22.

Problem: A metric space $\mathcal{M}$ is called separable if it contains a countable dense subset. Show that $\mathbb{R}^n$ is separable.

Proof: We prove the much, much deeper following theorem.

Lemma: Let $\left\{X_n\right\}_{n\in\mathbb{N}}$ be a countable class of non-empty separable topological spaces with corresponding countable dense subsets $\left\{\mathfrak{D}_n\right\}_{n\in\mathbb{N}}$. Then, $\displaystyle X=\prod_{n\in\mathbb{N}}X_n$ is separable with the product topology.

Proof: For each $n\in\mathbb{N}$ select an arbitrary but fixed $\mathfrak{d}_n\in\mathfrak{D}_n$ and define $\displaystyle \mathcal{D}_m=\prod_{n\in\mathbb{N}}D_n$ where

$D_n=\begin{cases} \mathfrak{D}_n\quad\text{if}\quad n\leqslant m \\ \{\mathfrak{d}_n\}\quad\text{if}\quad n>m\end{cases}$

Clearly we have that $D_m\simeq\mathfrak{D}_1\times\cdots\times\mathfrak{D}_m$ and thus countable (the finite product of countable sets is countable). So, let

$\displaystyle \mathfrak{D}=\bigcup_{m\in\mathbb{N}}\mathcal{D}_m$

Then $\mathfrak{D}$, being the countable union of countable sets, is countable. We now claim that it is dense in $X$. To see this let $\bold{x}\in X$ be arbitrary and $\bold{N}$ any basic neighborhood of it. Then, there are only finitely many indices $\{j_1,\cdots,j_k\}$ such that $\pi_{j_k}(\bold{N})\ne X_{j_k}$. So, let $J=\max\{j_1,\cdots,j_k\}$. Now, since for each $k=1,\cdots,J$ we have that $\pi_k(\bold{N})$ is a neighborhood of $\pi_k(\bold{x})$ in $X_k$ there is some $d_k\in\mathfrak{D}_k\cap\pi_k(\bold{N})$. So, doing this for $k=1,\cdots,J$ we arrive at $J$ points $d_1,\cdots,d_J$. So, then

$\displaystyle \prod_{n\in\mathbb{N}}\{x_n\},\quad x_n=\begin{cases}d_n\quad\text{if}\quad n\leqslant J\\ \mathfrak{d}_n\quad\text{if}\quad n>J\end{cases}\in D_J\cap\bold{N}$

Which proves that $\mathfrak{D}$ is dense. The conclusion follows. $\blacksquare$

The above is kind of heavy duty (though it works since the product topology and usual topology coincide). For people who are more interested in analysis, the real way to do this is to note that $\mathbb{Q}^n$ is countable and note that for any $(x_1,\cdots,x_n)\in\mathbb{R}^n$ you can choose corresponding $(q_1,\cdots,q_n)\in\mathbb{Q}^n$ such that $|x_k-q_k|<\sqrt{\frac{\varepsilon^2}{4n}}$ and do the simple calculation.

23.

Problem: A collection $\left\{V_\alpha\right\}_{\alpha\in\mathcal{A}}$ of open subsets of $\mathcal{M}$ ( a metric space) is said to be a base if for every $x\in\mathcal{M}$ and every neighborhood $U$ of it there is some $V_\alpha$ such that $x\in V_\alpha\subseteq U$. Prove that every separable metric space has a countable base (in other words that it’s second countable).

Proof: Let $\mathfrak{D}$ be the countable dense subset and let $\mathfrak{B}=\left\{B_q(\xi):q\in\mathbb{Q},\text{ }\xi\in\mathfrak{D}\right\}$. Clearly $\mathfrak{B}$ is countable . To see this we merely note that $\eta:\mathfrak{D}\times\mathbb{Q}\to\mathfrak{B}:(\xi,q)\mapsto B_q(\xi)$ is clearly a surjection. We shall now prove that $\mathfrak{B}$ (as it’s letter suggests) is a base. So, let $x\in\mathcal{M}$ be arbitrary and $\varepsilon>0$ be given. Since $\mathfrak{D}$ is dense in $\mathcal{M}$ there exists some $\xi \in B_{\frac{\varepsilon}{2}}(x)\cap\mathfrak{D}$. So, let $\delta=d(x,\xi)<\frac{\varepsilon}{2}$ and choose $q\in\mathbb{Q}$ such that $\delta. then, $x\in B_q(\xi)$ and if $y\in B_q(\xi)$ we have that $d(y,x)\leqslant d(y,\xi)+d(\xi,x) and thus $y\in B_q(\xi)$. The conclusion follows. $\blacksquare$

24.

Problem: Let $\mathcal{M}$ be a metric space in which every infinite subset of it has a limit point. Prove that $\mathcal{M}$ is separable.

Proof: Consider the open cover $\left\{B_\delta(x)\right\}_{x\in\mathcal{M}}$ and suppose that it did not have a finite subcover. Then, we may form a sequence $\left\{x_n\right\}_{n\in\mathbb{N}}$ such that $d(x_n,x_m)\geqslant \delta,\text{ }m, for otherwise if there existed some $n_0\in\mathbb{N}$ such that for no $x\in\mathcal{M}$ we have that $d(x,x_{n_0}),\cdots,d(x,x_1)\geqslant \delta$ then for every $x\in\mathcal{M}$ we have that

$d(x,x_1)<\delta\text{ or }\cdots\text{ or }d(x,x_{n_0-1})<\delta$

or in other words that $x\in B_{\delta}(x_1)\cup\cdots\cup B_{\delta}(x_{n_0-1})$ contradicting the assumption that $\left\{B_{\delta}(x)\right\}_{x\in\mathcal{M}}$ has no finite subcover. But, let $x\in\mathcal{M}$ and consider $B_{\delta}(x)$ clearly $B_{\delta}(x)$ contains at most one point of $\left\{x_n:n\in\mathbb{N}\right\}$. To see this suppose that it contained more than one point, say $x_m,x_n\in B_{\delta}(x)$ then we may assume WLOG that $m but this contradicts the construction of the sequence since $d(x_m,x_n)\geqslant \delta$. It follows that $\left\{B_{\delta}(x)\right\}_{x\in\mathcal{M}}$ has a finite subcover.  So, let for each $m\in\mathbb{N}$ we may cover $\mathcal{M}$ with finitely many open balls of the form $B_{\frac{1}{m}}(x)$. So, let $A_m$ be the finite set of points such that $B_{\frac{1}{m}}(x_1)\cup\cdots\cup B_{\frac{1}{m}}(x_n)=\mathcal{M}$ and let $\displaystyle A=\bigcup_{m=1}^{\infty}A_m$. Clearly $A$ is countable and given any $x\in\mathcal{N}$ and any neighborhood $U$ of it we may (by the Archimedean principle) find some $B_{\frac{1}{m}}(x)$ such that $B_{\frac{1}{m}}(x)\subseteq U$. But, since every point of $\mathcal{M}$ is less than $\frac{1}{m}$ in distance from some point of $A_m$ it follows there is some $\xi\in B_{\frac{1}{m}}\cap A_m$. The conclusion follows. $\blacksquare$

Remark: All we really showed above (the hard part anyways) is that every limit point compact metric space is totally bounded. As we prove in the next theorem limit point compactness is implied by compactness in metric spaces (they are in fact equivalent) and thus every compact metric space is totally bounded. It actually is true that any space such that every $\varepsilon>0$ there is a countable $\varepsilon$-net is separable, this is done precisely as in the last part of the above.

25.

Problem: Prove that every compact metric space $\mathcal{M}$ has a countable base.

Proof: It suffices to show that every infinite subset of $\mathcal{M}$ has a limit point, from where it will follow from 24. that it is separable, and thus second countable. So, let $E\subseteq\mathcal{M}$ have no limit point. Then, for each $x\in\mathcal{M}$ there exists some open ball $B_{\delta_x}(x)$ such that $E\cap B_{\delta_x}(x)\subseteq\{x\}$. So, we may cover $\mathcal{M}$ and so by assumption this open cover must admit a finite subcover $\left\{B_{\delta_{x_1}}(x_1),\cdots,B_{\delta_{x_n}}(x_n)\right\}$. So,

$E=E\cap\mathcal{M}=\left(E\cap B_{\delta_{x_1}}(x_1)\right)\cup\cdots\cup\left(E\cap B_{\delta_{x_n}}(x_n)\right)\subseteq\left\{x_1,\cdots,x_n\right\}$

The conclusion follows. $\blacksquare$

26.

Problem: Let $\mathcal{M}$ be a metric space such that every infinite subset has a limit point. Show that $\mathcal{M}$ is compact.

Proof:

Lemma: Let $\mathcal{M}$ be a limit point compact metric space and $\left\{F_n\right\}_{n\in\mathbb{N}}$ a sequence of descending ($F_n\supseteq F_{n+1}$) of non-empty closed subsets of $\mathcal{M}$. Then, $\displaystyle \bigcap_{n\in\mathbb{N}}F_n\ne\varnothing$

Proof: For each $n\in\mathbb{N}$ choose some $x_n\in F_n$. We break this into two cases,  namely whether the set of all these $x_n$‘s is finite or infinite.

Case 1: Suppose that $\left\{x_n:n\in\mathbb{N}\right\}$ is finite but $\displaystyle \bigcap_{n\in\mathbb{N}}F_n=\varnothing$. We first note that for each $x\in F_1$ there is some $N_x\in\mathbb{N}$ such that $N_x\leqslant n\implies x\notin U_n$. To see this assume not and let $n\in\mathbb{N}$ then $x\in F_n$ otherwise $x\notin F_n$ and thus $x\notin F_{n+1},\cdots$ which was assumed to be impossible.  So, $x\in F_n$ for all $n\in\mathbb{N}$ and thus $\displaystyle x\in\bigcap_{n\in\mathbb{N}}F_n$ which contradicts our assumption that it was empty. So, for each $x\in \left\{x_n:\in\mathbb{N}\right\}$ there exists some $N_x$ for which the above idea is true. Let $N=\max\left\{N_x:x\in\left\{x_n:n\in\mathbb{N}\right\}\right\}$ then $\left\{x_n:n\in\mathbb{N}\right\}\cap F_{N+1}=\varnothing$ which contradicts the construction of the set. It follows that the intersection $\displaystyle \bigcap_{n\in\mathbb{N}}F_n$ is non-empty.

Case 2: Suppose that $\left\{x_n:n\in\mathbb{N}\right\}$ is infinite. By assumption it then has a limit point $\xi$. So, we show that $\displaystyle \xi\in\bigcap_{n\in\mathbb{N}}F_n$ by showing that it is contained in a fixed but arbitrary $F_{n_0}$. So, suppose that $\xi\notin F_{n_0}$ then $\xi\in \mathcal{M}-F_{n_0}$ which is open. But, since $F_{n_0}\supseteq F_{n_0+1}\cdots$ it follows that $\left(\mathcal{M}-F_{n_0}\right)\cap\left\{x_n:n\in\mathbb{N}\right\}$ is finite contradicting that $\xi$ is a limit point of that set. It follows that $\displaystyle \xi\in\bigcap_{n\in\mathbb{N}}F_n$. $\blacksquare$

So, now let $\Omega$ be an open cover for $\mathcal{M}$. Since $\mathcal{M}$ is second countable we know that $\Omega$ admits a countable subcover $\left\{U_n\right\}_{n\in\mathbb{N}}$. Now, suppose that $U_1\cup\cdots\cup U_m\ne\mathcal{M}$ for every $m\in\mathbb{N}$ then $\left\{F_n\right\}_{n\in\mathbb{N}}$ where $F_n=\mathcal{M}-\left(U_1\cup\cdots\cup U_n\right)$ is a sequence of decreasing, non-empty closed sets. But,

$\displaystyle \bigcap_{n\in\mathbb{N}}F_n=\mathcal{M}-\bigcup_{n\in\mathbb{N}}\left(U_1\cup\cdots\cup U_n\right)=\mathcal{M}-\mathcal{M}=\varnothing$

Which contradicts the lemma. The conclusion follows. $\blacksquare$

27.

Problem: Let $\mathcal{M}$ be a metric space and call $\xi\in\mathcal{M}$ a condensation point of a set $E\subseteq\mathcal{M}$ if every neighborhood of $\xi$ contains uncountably many points of $E$. Now, suppose that $E\subseteq\mathbb{R}^n$ is uncountable and let $\mathfrak{C}$ be the set of all condensation points of $E$. a) Prove that $\mathcal{C}$ is perfect and show that $\left(\mathbb{R}-\mathfrak{C}\right)\cap E$ is countable.

Proof:

a) Let $x\in \mathfrak{C}$ but suppose that $x\notin D(\mathfrak{C})$ then there exists some neighborhood $U-\{x\}$ of it such that for every $y\in U$ there exists a neighborhood $N_y$ of it such that $\text{card }N_y\cap E\leqslant\aleph_0$. So, we now need a quick lemma

Lemma: Let $X$ be a second countable topological space and $U$ any subset of $X$. Then, if $\left\{V_\alpha\right\}_{\alpha\in\mathcal{A}}$ is a collection of open sets in $X$ such that $\displaystyle U\subseteq\bigcup_{\alpha\in\mathcal{A}}V_\alpha$ then there exists some countable subcollection $\left\{V_{\alpha_n}\right\}_{n\in\mathbb{N}}$ such that $\displaystyle U\subseteq\bigcup_{n\in\mathbb{N}}V_{\alpha_n}$

Proof: For each $u\in U$ there is some $V_\alpha$ such that $u\in V_\alpha$ but if $\mathfrak{B}$ is the countable base there is some $B\in\mathfrak{B}$ such that $u\in B\subseteq V_{\alpha}$. Doing this for each $u\in U$ clearly gives a collection $\left\{B_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\mathfrak{B}$ such that $\displaystyle U\subseteq\bigcup_{\beta\in\mathcal{B}}B_\beta$. But, by design it is clear that $\mathcal{B}$ is countable and so for each $\beta\in\mathcal{B}$ we may choose some (since this is how they were begotten) some $V_\beta\in\left\{V_\alpha\right\}_{\alpha\in\mathcal{A}}$ such that $B_\beta\subseteq V_\beta$. So, doing this for each $\beta\in\mathcal{B}$ we arrive at a countable collection $\left\{V_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\left\{V_\alpha\right\}_{\alpha\in\mathcal{A}}$ such that

$\displaystyle U\subseteq\bigcup_{\beta\in\mathcal{B}}B_\beta\subseteq\bigcup_{\beta\in\mathcal{B}}V_\beta$

from where the conclusion follows. $\blacksquare$

So, note that since $\mathbb{R}^n$ is second countable and we clearly have that $\left\{N_y\cap Y\right\}_{y\in U}$ is a collection of open sets whose union contains $U$ and so by the above lemma it follows that it must admit a countable subcover $\left\{N_{y_n}\cap U\right\}_{n\in\mathbb{N}}$ so then

$\displaystyle \text{card }U\cap E=\text{card }\bigcup_{n\in\mathbb{N}}\left(U\cap N_{y_n}\right)=\text{card }\bigcup_{n\in\mathbb{N}}\left(U\cap N_{y_n}\cap E\right)\leqslant \aleph_0$

which contradicts that $x\in \mathfrak{C}$. So, this proves that $\mathfrak{C}\subseteq D(\mathfrak{C})$. Conversely, let $x\in D(\mathfrak{C})$ then every neighborhood $U$ of $x$ contains a point of $\mathfrak{C}$ , but since $U$ is also a neighborhood of that point it follows that $\text{card }U\cap E>\aleph_0$ from where it follows by the arbitrariness of $U$ that $x\in \mathfrak{C}$.

b) Let $\mathcal{W}=\left\{B\in\mathfrak{B}:\text{card }B\cap E\leqslant \aleph_0\right\}$. We claim that $\displaystyle \Sigma\overset{\text{def}}{=}\bigcup_{B\in\mathcal{W}}B=\mathbb{R}-\mathfrak{C}$. To see this let $x\in \Sigma$ then there is some $B\in\mathfrak{B}$ such that $\text{card }B\cap E\leqslant \aleph_0$ and thus $x\notin \mathfrak{C}$. Conversely, let $x\in\mathbb{R}-\mathfrak{C}$ then there exists some neighborhood $U$ of $x$ such that $\text{card }U\cap E\leqslant\aleph_0$ but there exists some $B\in\mathfrak{B}$ such that $x\in B\subseteq U$. Thus, $x\in\Sigma$. So, we see that

$\displaystyle \text{card }E\cap\Sigma=\text{card }E\cap\bigcup_{B\in\mathcal{W}}B=\text{card }\bigcup_{B\in\mathcal{W}}\left(B\cap E\right)\leqslant\aleph_0$

The last part gotten from the fact that $\mathcal{W}\subseteq\mathfrak{B}$ is countable and each $E\cap B$ also countable. The conclusion follows. $\blacksquare$