# Abstract Nonsense

## Algebraic Numbers Are Countable

Point of post: In this post I will give a proof that the algebraic numbers are countable, and thus we will prove that there are non-algebraic (transcendental) numbers.

Remark: Personally I find my proof much simpler and more intuitive than the usual method by considering the “height” of numbers (viz. Rudin, pg. 43)

October 21, 2010

## 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. $\blacksquare$

May 14, 2010

## 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. $\blacksquare$

May 14, 2010

## 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. $\blacksquare$

May 14, 2010

## 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.

May 13, 2010

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

Just for fun I’ve decided to do the entire section on topology in Rudin’s Principles of Mathematical Analysis. The problems themselves aren’t particularly difficult, but some are interesting and it will give me a chance to do some refreshing. Also, a friend of mine is starting Rudin and it will be nice to have done them in case he has any questions. I will probably prove more general results from topology if I feel like it. So…

1.

Problem: Prove that the empty set is a subset of every set.

Proof: Suppose that we are working in some universal set $U$ and there existed some $E\subseteq U$ such that $\varnothing\not\subseteq E$. Then, by definition there exists some $x\in \varnothing$ such that $x\notin E$. But, this is clearly preposterous since the statement $x\in \varnothing$ is never true. $\blacksquare$

2.

Problem: A complex number $z$ is said to be algebraic if there are integers $a_1,\cdots,a_n$, not all zero, such that

$a_1+\cdots+a_nz^n=0$

Let $\mathbb{A}$ denote the set of all algebraic numbers. Prove that $\text{card }\mathbb{A}=\aleph_0$.

Proof: Let $\Omega_n(a_1,\cdots,a_n)=\left\{z\in\mathbb{C}:a_1+\cdots+a_nz^n=0\right\}$. By the fundamental theorem of algebra we have that $\text{card }\Omega_n(a_1,\cdots,a_n)\leqslant n$. So, clearly $\displaystyle \bigcup_{(a_1,\cdots,a_n)\in\mathbb{Z}^n}\Omega_n(a_1,\cdots,a_n)$ being the countable union of countable sets is countable. Thus, we claim that

$\displaystyle \mathbb{A}\subseteq\bigcup_{n\in\mathbb{N}}\bigcup_{(a_1,\cdots,a_n)\in\mathbb{Z}^n}\Omega_n(a_1,\cdots,a_n)$

from where it will follows that $\mathbb{A}$ being a subset of a countable union of countable sets is countable. So, let $z\in\mathbb{A}$ then there exists $a_1,\cdots,a_m\in\mathbb{Z}^m$ such that $a_1+\cdots+a_mz^m=0$. Thus, clearly $z\in\Omega_m(a_1,\cdots,a_m)\subseteq$ from where the conclusion follows. $\blacksquare$

3.

Problem: Prove that there exists real numbers which aren’t algebraic.

Proof: Clearly since $\mathbb{A}$ is countable so is $\mathbb{R}\cap\mathbb{A}=\mathbb{R}_{\mathbb{A}}$. And thus, if $\mathbb{R}-\mathbb{R}_{\mathbb{A}}$ were countable then so would $\mathbb{R}_{\mathbb{A}}\cup\left(\mathbb{R}-\mathbb{R}_{\mathbb{A}}\right)=\mathbb{R}$ which is of course absurd. It follows that there are uncountably many non-algebraic (transcendental) numbers. $\blacksquare$

4.

Problem: Is the set of all irrational numbers countable?

Proof: Of course not. For using a similar argument to the above, if $\mathbb{I}=\mathbb{R}-\mathbb{Q}$ were countable then  $\mathbb{Q}\cup\mathbb{I}=\mathbb{R}$ would be as well. $\blacksquare$

5.

Problem: Construct a bounded set of real numbers with exactly three limit points.

Proof: Why not do it for an arbitrary natural number. Let $\displaystyle E_0=\{0\}\cup\left\{\frac{1}{n}:n\in\mathbb{N}\right\}$. Evidently $E_0$ has precisely one limit point. Now, let $E_m=\left\{e+m:e\in E_0\right\}$ then it is equally clear $E_m$ has exactly one limit point (namely $m$). So, the set $K_{m}=E_0\cup\cdots\cup E_{m-1}$ has precisely $m$ limit points as one can easily check. $\blacksquare$

6.

Problem: Let $D(E)$ be the set of all limit points of a set $E\subseteq\mathcal{M}$ (where $\mathcal{M}$ is a metric space). Prove that a) $D(E)$ is closed , b) $D(E)=D\left(\overline{E}\right)$, c) Do $E$ and $D(E)$ always have the same limit points?

Proof:

a) This is true in any $T_1$ topological space. But, first we need a quick lemma

Lemma: Let $X$ be a $T_1$ space, and $U\subseteq X$. Then, if $x\in D(U)$ then for very neighborhood $V$ of $x$ we have that $V\cap U$ is infinite.

Proof: Suppose not, and there existed a neighborhood $V$ of $x$ such that $V\cap U=\{u_1,\cdots,u_m\}$. Since $X$ is $T_1$ for each $u_k$ there exists some neighborhood $N_k$ of $x$ such that $u_k\notin N_k$. Thus, $V\cap N_1\cap\cdots\cap N_m$ is a neighborhood of $x$ which does not intersect $U$, which of course contradicts that $x\in D(U)$. $\blacksquare$

So, using this let $X$ be a $T_1$ space, then for any $E\subseteq X$ it is true that $D(E)$ is closed. To see this let $x\in D(D(E))$ and let $N$ be any neighborhood of it. By assumption there exists some $y\in N\cap D(E)$ and thus since $N$ is a neighborhood of $y$ also there exists infinitely many points of $E$ in $N$. In particular there exists some $e\in N\cap E$ such that $e\ne x$. Since $N$ was arbitrary it follows that $x\in D(E)$. The conclusion follows.

b) Let $x\in D(E)$ then every neighborhood $N$ contains a point of $E$ different from itself, and thus a point of $\overline{E}$ different from itself. Conversely, let $x\in D(\overline{E})$  and let $N$ be any neighborhood of it. We must have that there is some $y\in N\cap \overline{E}$. Now, if $y\in E$ we’re done, and if not by the previous lemma since $y$ is a limit point of $E$ it must be that $N$ contains infinitely many points of $E$ anyways, and thus a point of $E$ different from $x$. The conclusion follows.

c) Of course not. $\left\{0\right\}\cup\left\{\frac{1}{n}:n\in\mathbb{N}\right\}\subseteq\mathbb{R}$ with the usual topology has exactly one limit point, namely $0$. But, a singleton in Euclidean space does not have an limit points.

7.

Problem Let $\mathcal{M}$ be a metric space and $E_1,\cdots,\subseteq\mathcal{M}$. a) Show that $\displaystyle \overline{\bigcup_{m=1}^{n}E_m}=\bigcup_{m=1}^{n}\overline{E_m}$.  b) show that example that $\displaystyle \overline{\bigcup_{m=1}^{\infty}E_m}\supseteq\bigcup_{m=1}^{\infty}\overline{E_m}$ and the inclusion can be strict.

Proof:

a) This is true in any topological space $X$. Suppose $x\notin\overline{E_1\cup\cdots\cup E_n}$ then there exists a neighborhood $N$ of $x$ such that $N\cap\left(E_1\cup\cdots\cup E_n\right)=(N\cap E_1)\cup\cdots\cup (N\cap E_n)=\varnothing$ and thus $N\cap E_k=\varnothing,\text{ }k=1,\cdots,n$ and thus $x\notin \overline{E_k},\text{ }k=1,\cdots,n$ and so $x\notin\overline{E_1}\cup\cdots\cup\overline{E_n}$. Conversely, let $x\notin\overline{E_1}\cup\cdots\cup\overline{E_n}$. Then, since $x\notin \overline{E_k},\text{ }k=1,\cdots,n$ there exists some neighborhood $U_k$ of it such that $U_k\cap E_k=\varnothing$. Thus, clearly $U_1\cap\cdots\cap U_n$ is a neighborhood of $x$ which does not intersect $E_1\cup\cdots\cup E_n$ and thus $x\notin\overline{E_1\cup\cdots\cup E_n}$. The conclusion follows.

b) Let $\displaystyle x\notin\overline{\bigcup_{m=1}^{\infty}E_m}$ then there exists a neighborhood $N$ of $x$ such that $\displaystyle N\cap\bigcup_{m=1}^{\infty}E_m=\bigcup_{m=1}^{\infty}\left(N\cap E_m\right)=\varnothing$ and thus $N\cap E_m=\varnothing,\text{ }m\in\mathbb{N}$. It follows that $x\notin \overline{E_m},\text{ }m\in\mathbb{N}$ and thus $\displaystyle x\notin\bigcup_{m=1}^{\infty}\overline{E_m}$. But, as was noted the inclusion can be strict. Give $\mathbb{R}$ the usual topology. Then, for each $q\in\mathbb{Q}$ we have the singleton $\{q\}$ is closed and thus $\overline{\{q\}}=\{q\}$. So,

$\displaystyle \bigcup_{q\in\mathbb{Q}}\overline{\{q\}}=\bigcup_{q\in\mathbb{Q}}\{q\}=\mathbb{Q}\subsetneq\overline{\bigcup_{q\in\mathbb{Q}}\{q\}}=\overline{\mathbb{Q}}=\mathbb{R}$

8.

Problem: a) Is every point of every open set $E\subseteq\mathbb{R}^2$ a limit point of $E$? Answer the same for closed sets in $\mathbb{R}$?

Proof:

a) Of course, I mean this is true in any topological space. If $E$ is open then each point of $E$ is an interior point and thus trivially a limit point.

b) No, take a singleton.

9.

Problem: Let $E^{\circ}$ denote the set of all interior points for $E$ in a metric space $\mathcal{M}$. Prove that:

a) $E^{\circ}$ is always open

b) $E^{\circ}$ is open if and only if $E^{\circ}=E$.

c) If $G\subseteq E$ and $G$ is open then $G\subseteq E^{\circ}$.

d) $\mathcal{M}-E^{\circ}=\overline{\mathcal{M}-E}$

e) Do $E$ and $\overline{E}$ always have the same interiors?

f) Do $E$ and $E$ always have the same interiors?

Proof:

a) Let $x\in E^{\circ}$ then there exists a neighborhood $N$ of $x$ such that $N\subseteq E$, but for each $y\in N$ we also have that $N$ is a neighborhood of it which is contained in $E$. Thus, $N\subseteq E^{\circ}$.

b) If $E=E^{\circ}$ the above proves that $E$ is open. Conversely, since a set in a metric space is open if and only if each point is an interior point the converse is trivial.

c) Let $x\in G$ then $G$ is automatically a neighborhood of $x$ which is contained in $E$ so that $x\in E^{\circ}$.

d) Let $x\in\mathcal{M}-E^{\circ}$ then for every neighborhood $N$ of $x$ we have that $N\not\subseteq E$ and so $N\cap \mathcal{M}-E\ne\varnothing$ and thus $x\in\overline{\mathcal{M}-E}$. Conversely, let $x\in\overline{\mathcal{M}-E}$ then for every neighborhood $N$ of $x$ we have that $N\cap\mathcal{M}-E\ne\varnothing$ and thus $N\not\subseteq E$ and so $x\notin E^{\circ}$ or $x\in\mathcal{M}-E^{\circ}$.

e) Of course not. It is trivial to check that if one gives $\mathbb{R}$ the usual topology then $\mathbb{Q}^{\circ}=\varnothing$ (since it’s complement is dense in $\mathbb{R}$) and $\overline{\mathbb{Q}}=\mathbb{R}$.

f) The above works as a counterexample to this as well.

10.

Problem: Let $\mathcal{M}$ be an infinite set and define $d:\mathcal{M}\times\mathcal{M}\to\mathbb{R}$ by

$d(x,y)=\begin{cases} 0 \quad\text{if}\quad x=y \\ 1\quad\text{if}\quad x\ne y \end{cases}$

a) Prove that this is a metric. b) Which subsets of the resulting metric space are open? c) Which are closed? d) Which are compact?

Proof:

a) Clearly $d(x,y)\geqslant 0$ and $d(x,y)=0\Leftrightarrow x=y$. Also, since equality and inequality are transitive symmetry immediately follows. The triangle inequality is just a set of three cases. If $x=y=z$ then $d(x,y)=0\leqslant d(x,z)+d(y,z)=0+0$. If $x=z,y\ne z$ then $d(x,y)=1\leqslant d(x,z)+d(y,z)=1+0=1$. If $x\ne z,y\ne z$ then $d(x,y)\leqslant 1\leqslant d(x,z)+d(y,z)=1+1$.

b) For every $x\in\mathcal{M}$ we have that $\{x\}$ is open since $B_{\frac{1}{2}}(x)=\{x\}$. Thus, all subsets of $\mathcal{M}$ being the union of singletons are open.

c) From b) it readily follows that every subset of $\mathcal{M}$ is also open.

d) $E\subseteq\mathcal{M}$ will be compact if and only if $E$ is finite. To see this suppose that $E$ is infinite, then $\left\{\{e\}\right\}_{e\in E}$ is evidently an open cover of it which admits no finite subcover. The converse is trivial.

11.

Problem: Let the following be maps from $\mathbb{R}^2$ to $\mathbb{R}$, and determine if they determine a metric. a) $d(x,y)=(x-y)^2$, b) $d(x,y)=\sqrt{|x-y|}$, c) $d(x,y)=|x^2-y^2|$, d) $d(x,y)=|x-2y|$, e) $\displaystyle d(x,y)=\frac{|x-y|}{1+|x-y|}$

Proof:

a) This is not a metric. $(4-1)^2=9>5=4+1=(4-2)^2+(1-2)^2$

b) This is a metric as one (I’m to lazy to tex it) can check.

c) Nope, $d(1,-1)=0$.

d) Negatory, $d(1,\tfrac{1}{2})=0$.

e) Yes. In fact, (once again I omit the proof) if $d$ is a metric so is $\displaystyle \frac{d}{1+d}$

12.

Problem: Let $K\subseteq\mathbb{R}$ with the usual topology be defined by $K=\{0\}\cup\left\{\frac{1}{n}:n\in\mathbb{N}\right\}$. Prove directly (i.e. no Heine-Borel) that $K$ is compact.

Proof: This is true in any topological space $X$. But, for the sake of convenience let us do it merely in a Hausdorff space $X$. So, let $\{x_n\}_{n\in\mathbb{N}}$ is convergent sequence in $X$ with $x_n\to x$. Then, $C=\{x\}\cup\left\{x_n:n\in\mathbb{N}\right\}$ is a compact subspace of $X$.

Proof: Let $\Omega$ be an open cover for $C$. Choose $U\in\Omega$ such that $x\in U$. Then, since $x_n\to x$ all but finitely may values of $\{x_n:n\in\mathbb{N}\}$ lie in $U$. So, for each of the other finitely many points of $C-U$ choose any element of $\Omega$ which contains them. Clearly this will be a finite subcover. $\blacksquare$

May 13, 2010