# Abstract Nonsense

## Miscellaneous Rudin Like Topology questions

Problem: Let $X$ be a second countable topological space and $U\subseteq X$ be such that $\text{card }U>\aleph_0$. Then, $\text{card }U\cap D(U)>\aleph_0$

Proof: Suppose that $D(U)\cap U$ is countable then $U-(U\cap D(U))=U-D(U)$ is uncountable. But, by definition for each $x\in U-D(U)$ there exists some neighborhood $V_x$ of it such that $U\cap V_x=\{x\}$. So, clearly $\left\{V_x\cap (U-D(U))\right\}_{x\in U-D(U)}$ is an open cover for $U-D(U)$ and since $X$ is second countable Lindelof’s theorem guarantees that it must have a countable subcover $\left\{V_{x_n}\cap (U-D(U))\right\}_{n\in\mathbb{N}}$. Thus,

$\displaystyle U-D(U)=\bigcup_{n\in\mathbb{N}}\left(V_{x_n}\cap(U-D(U))\right)\subseteq\bigcup_{n\in\mathbb{N}}\left(V_{x_n}\cap U\right)=\left\{x_n:n\in\mathbb{N}\right\}$

But this contradicts that $U-D(U)$ is uncountable. The conclusion follows. $\blacksquare$

Corollary: If $f:\mathbb{C}\to\mathbb{C}$ and $\mathcal{S}$ is the set of all isolated singularities then $\text{card }\mathcal{S}\leqslant\aleph_0$

Corollary: Let $\mathcal{M}$ be a separable metric space and $E\subseteq\mathcal{M}$ uncountable, then there exists a sequence $\{e_n\}_{n\in\mathbb{N}}$ in $E$ such that $e_n\to e\in E$.

Problem: Let $\mathcal{M}$ be a metric space and $E\subseteq \mathcal{M}$ perfect and $U\subseteq\mathcal{M}$ open. Then, $\overline{E\cap U}$ is perfect.

Proof: The fact that $D(\overline{E\cap U})\subseteq \overline{E\cap U}$  follows from the fact that it’s closed. Now, let $x\in \overline{E\cap U}$ and let $V$ be any neighborhood of it. Then, by definition we have that $V\cap (E\cap U)\ne \varnothing$ and so there exists some $y\in V\cap E\cap U$. Now, since $y\in E$ and $V\cap U$ is a neighborhood of it it follows that there are infinitely many values of $E$ in it. In other words $V\cap (U\cap E)$ is infinite and in particular there exists some $y\in V\cap (U\cap E)$ such that $y\ne x$. Thus, $y\in D(\overline{E\cap U})$. The conclusion follows. $\blacksquare$

Corollary: If $\mathcal{M}$ is a metric space and $\varnothing\subsetneq E\subseteq\mathcal{M}$ is perfect we can choose $x\in E$ and see that $\overline{E\cap B_{1}(x)}$ and  we have a perfect subset of $\mathcal{M}$ which is also bounded.

Problem: Let $\mathcal{M}$ be a compact metric space. Prove that $\mathcal{M}$ is connected if and only if it cannot be written as a union $\mathcal{M}=A\cup B$ with $d(A,B)=\inf\left\{d(a,b):a\in A,b\in B\right\}>0$

Proof: It is tacitly assumed that $A,B\ne\varnothing$. So, in any metric space suppose that $\mathcal{M}=A\cup B$ with $d(A,B)>0$. Remember that $\overline{A}=\left\{x\in\mathcal{M}:d(x,A)=0\right\}$ and thus if $x\in B$ we have that $d(x,A)\geqslant d(A,B)>0$ and thus $x\notin \overline{A}$. Thus, $\overline{A}\cap B=A\cap\overline{B}=\varnothing$ (the reverse direction gotten using the exact same logic) from where it follows that $\mathcal{M}$ is disconnected.

Before we continue to prove the converse we prove a nice little lemma

Lemma: Let $\mathcal{M}$ be a compact metric space and $E,G\subseteq\mathcal{M}$ non-empty disjoint closed subspaces of $\mathcal{M}$. Then, $d(E,G)>0$.

Proof: Since $\mathcal{M}$ is compact and $E,G$ closed subspaces it follows that they themselves are compact and thus so is $E\times G$. So, notice that $F:E\times G\to\mathbb{R}:(e,g)\mapsto d(e,g)$ is continuous since it is the distance function (trivially continuous) restricted to $E\times G$. Thus, by the extreme value theorem it follows that $\inf\text{ }F(E\times G)=d(e_0,g_0)$ for some $(e_0,g_0)\in E\times G$ and since $e_0\ne g_0$ we see that $d(E,G)=d(e_0,g_0)>0$. $\blacksquare$

So, now assume that $\mathcal{M}$ were not connected. Then, $\mathcal{M}=E\cup G$ where $E,G$ are non-empty disjoint closed subsets of $\mathcal{M}$. But, by the lemma this contradicts the assumption that $\mathcal{M}$ cannot be written as $A\cup B$ with $d(A,B)>0$. $\blacksquare$

Problem: Let $\mathcal{M}$ be a separable metric space, then for any subspace $\mathcal{N}$ we have that $\mathcal{N}$ is separable.

Proof: We prove something stronger. But, first we prove a small lemma

Lemma: Let $X$ be a second countable topological space, then any subspace $Y$ is second countable.

Proof: Let $\mathfrak{B}$ be the countable base for $X$ and let $\mathfrak{B}_Y=\left\{B\cap Y:B\in\mathfrak{B}\right\}$. Clearly $\mathfrak{B}_Y$ is a countable collection of open subsets of $Y$. Now, let $y\in Y$ be arbitrary and let $U$ be any neighborhood of it. Then, by definition $U=V\cap Y$ for some open set $V$ in $X$. So, we may find some $B\in\mathfrak{B}$ such that $y\in B\subseteq V$ and thus $y\in B\cap Y\subseteq V\cap Y=U$ and since $B\cap Y\in\mathfrak{B}_Y$ the conclusion follows. $\blacksquare$

So, if $\mathcal{M}$ is a separable metric space, then as proven earlier it is second countable and thus so is (by the lemma) $\mathcal{N}$ but since every second countable space is trivially separable the conclusion follows. $\blacksquare$

Problem: Let $\mathcal{M}$ be a compact metric space and $\varphi:\mathcal{M}\to\mathcal{M}$ an isometry. Then, $\varphi$ is surjective.

Proof: Suppose $x\in\mathcal{M}-f(\mathcal{M})$ and consider $\left\{f^n(x)\right\}_{n\in\mathbb{N}}$. Since $\mathcal{M}$ is a compact metric space this must have a convergent subsequence $\left\{f^{n_m}(x)\right\}_{m\in\mathbb{N}}$. In particular, that sequence must be Cauchy. So, noting that $\{x\}f(\mathcal{M})$ are disjoint non-empty compact subspaces of a compact space it follows by an earlier lemma that $d(x,M)=\delta>0$4. So, since $\left\{f^{n_m}(x)\right\}_{m\in\mathbb{N}}$ is Cauchy there exists $M\in\mathbb{N}$ such that $d(f^{n_{M+1}}(x),f^{n_M}(x))<\delta$. So,

$\delta> d\left(f^{n_{M+1}}(x),f^{n_M}(x)\right)=d\left(f^{n_M}(x),f^{n_{M-1}}(x)\right)=\cdots=d(f(x),x)>\delta$

which is a contradiction. $\blacksquare$

Problem: Let $\mathcal{M}$ be a compact metric space and $\varphi:\mathcal{M}\to\mathcal{M}$ be a function such that

$d(\varphi(x),\varphi(y))

for all distinct $x,y\in\mathcal{M}$. Prove that $\varphi$ has precisely one fixed point.

Proof: Clearly if there is a fixed point it’s unique because if $\varphi(x)=x,\varphi(y)=y,\text{ }x\ne y$ then $d(x,y)=d(\varphi(x),\varphi(y)). Now, to see that it must have a fixed point we note that

$\psi:\mathcal{M}\to\mathcal{M}:x\mapsto d(x,\varphi(x))$

is continuous since evidently the diagram



commutes. So, since $\psi$ is continuous and $\mathcal{M}$ it follows that $\psi$ obtains a minimum at some $x_0\in\mathcal{M}$. Now, assume that $\varphi(x_0)\ne x_0$ then $d(\varphi(\varphi(x_0)),\varphi(x_0)) contradicting that $\psi$ achieved it’s minimum at $x_0$. The conclusion follows. $\blacksquare$