# Abstract Nonsense

## Munkres Chapter 2 Section 19 (Part II)

9.

Problem: Show that the axiom of choice (AOC) is equivalent to the statement that for any indexed family $\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}}$ of non-empty sets, with $\mathcal{A}\ne\varnothing$, the product $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing$

Proof: This is pretty immediate when one writes down the actual definition of the product, namely:

$\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha=\left\{\bold{x}:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}:\bold{x}(\alpha)\in U_\alpha,\text{ for every }\alpha\in\mathcal{A}\right\}$

So, if one assumes the AOC then one must assume the existence of a choice function

$\displaystyle c:\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\to\bigcup_{\alpha\in\mathcal{A}},\text{ such that }c\left(U_\alpha\right)\in U_\alpha\text{ for all }\alpha\in\mathcal{A}$

So, then if we consider $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\overset{\text{def.}}{=}\Omega$ as just a class of sets, the fact that we have indexed them implies there exists a surjective “indexing function$$i:\mathcal{A}\to\Omega$ where clearly since we have already indexed out set we have that $i:\alpha\mapsto U_\alpha$. So, consider $c\circ i:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}}U_\alpha$ This is clearly a well-defined mapping and $\left(c\circ i\right)(\alpha)=c\left(U_\alpha\right)\in U_\alpha$ and thus $\displaystyle c\circ i\in\prod_{\alpha\in\mathcal{A}}U_\alpha$ from where it follows that $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing$ Conversely, let $\Omega$ be a class of sets and let $i:\mathcal{A}\to\Omega$ be an indexing function. We may then index $\Omega$ by $\Omega=\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}}$. Then, by assumption $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing$ Thus there exists some $\displaystyle \bold{x}\in\prod_{\alpha\in\mathcal{A}}U_\alpha$ Such that $\displaystyle \bold{x}:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}},\text{ such that }\bold{x}(\alpha)\in U_\alpha$ Thus, we have that $\displaystyle \bold{x}\circ i^{-1}:\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\to\bigcup_{\alpha\in\mathcal{A}}$ is a well-defined mapping with $\displaystyle \bold{x}\left(U_\alpha\right)\in U_\alpha$ For each $\alpha\in\mathcal{A}$. It follows that we have produced a choice function for $\Omega$ and the conclusion follows. $\blacksquare$ Remark: We have assumed the existence of a bijective indexing function $i:\mathcal{A}\to\Omega$, but this is either A) a matter for descriptive set theory or B) obvious since $\text{id}:\Omega\to\Omega$ satisfies the conditions. This depends on your level of rigor. 10. Problem: Let $A$ be a set; let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ be an indexed family of spaces; and let $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ be an indexed family of functions $f_\alpha:A\to X_\alpha$ a) Prove there is a unique coarsest topology $\mathfrak{J}$ on $A$ relative to whish each of the functions $f_\alpha$ is continuous. b) Let $\mathcal{S}_\beta=\left\{f_\beta^{-1}\left(U_\beta\right):U_\beta\text{ is ope}\text{n in}X_\beta\right\}$ and let $\displaystyle \mathcal{S}=\bigcup_{\alpha\in\mathcal{A}}$. Prove that $\mathcal{S}$ is a subbasis for $\mathfrak{J}$. c) Show that the map $g:Y\to A$ is continuous relative to $\mathfrak{J}$ if and only if each map $f_\alpha\circ g:Y\to X_\alpha$ is continuous. d) Let $\displaystyle f:A\to\prod_{\alpha\in\mathcal{A}}X_\alpha$ be defined by the equation $f(x)=\left(f_\alpha(x)\right)_{\alpha\in\mathcal{A}}$ Let $Z$ denote the subspace of $f\left(A\right)$ of the product space $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$. Prove taht the image under $f$ of each element of $\mathfrak{J}$ is an open set in $Z$. Proof: a) We first prove a lemma Lemma: Let $\mathfrak{J}$ be a topology on $A$, then all the mappings $f_\alpha:A\to X_\alpha$ are continuous if and only if $\mathcal{S}\subseteq\mathfrak{J}$ where $\mathcal{S}$ is defined in part b). Proof:Suppose that all the mappings $f_\alpha:A\to X_\alpha$ are continuous. Then, given any open set $U_\alpha\in X_\alpha$ we have that $f_\alpha$ is continuous and so $f_\alpha^{-1}\left(U_\alpha\right)$ is open and thus $f_{\alpha}^{-1}\left(U_\alpha\right)\in\mathfrak{J}$ from where it follows that $\mathcal{S}\subseteq\mathfrak{J}$. Conversely, suppose that $\mathcal{S}\subseteq\mathfrak{J}$. It suffices to prove that $f_\alpha:A\to X_\alpha$ for a fixed but arbitrary $\alpha\in\mathcal{A}$. So, to do this let $U$ be open in $X_\alpha$ then $f_{\alpha}^{-1}\left(U\right)\in\mathfrak{J}$ and thus by assumption $f_\alpha^{-1}\left(U\right)\in\mathfrak{J}$; but this precisely says that $f_\alpha^{-1}\left(U\right)$ is open in $A$. By prior comment the conclusion follows. $\blacksquare$ So, let $\mathcal{C}=\left\{\mathfrak{T}:\mathfrak{J}\text{ is a topology on }A\text{ and }\mathcal{S}\subseteq\mathfrak{J}\right\}$ and let $\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{I}\in\mathcal{C}}\mathfrak{T}$ By previous problem $\mathfrak{J}$ is in fact a topology on $A$, and by our lemma we also know that all the mappings $f_\alpha:A\to X_\alpha$ are continuous since $\mathcal{S}\subseteq\mathfrak{J}$. To see that it’s the coarsest such topology let $\mathfrak{U}$ be a topology for which all of the $f_\alpha:A\to X_\alpha$ are continuous. Then, by the other part of our lemma we know that $\mathcal{S}\subseteq\mathfrak{U}$ and thus $\mathfrak{U}\in\mathcal{C}$. So, $\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\mathcal{C}}\mathfrak{T}\subseteq\mathfrak{U}$ And thus $\mathfrak{J}$ is coarser than $\mathfrak{U}$. The uniqueness is immediate. b) It follows from the previous problem that we must merely show that $\mathcal{S}$ is a subbasis for the topology $\mathfrak{J}$. The conclusion will follow from the following lemma (which was actually an earlier problem, but we reprove here for referential reasons): Lemma: Let $X$ be a set and $\Omega$ be a subbasis for a topology on $X$. Then, the topology generated by $\Omega$ equals the intersection of all topologies which contain $\Omega$. Proof: Let $\mathcal{C}\left\{\mathfrak{T}:\mathfrak{T}\text{ is a topology on }X\text{ and }\Omega\subseteq\mathfrak{T}\right\}$ and $\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\mathcal{C}}\mathfrak{T}$ Also, let $\mathfrak{G}$ be the topology generated by the subbasis $\Omega$. Clearly since $\Omega\subseteq\mathfrak{G}$ we have that $\mathfrak{J}\subseteq\mathfrak{G}$. Conversely, let $U\in\mathfrak{G}$. Then, by definition to show that $U\in\mathfrak{J}$ it suffices to show that $U\in\mathfrak{T}$ for a fixed but arbitrary $\mathfrak{T}\in\mathcal{C}$. To do this we first note that by definition that $\displaystyle U=\bigcup_{\alpha\in\mathcal{A}}U_\alpha$ where each $U_\alpha=O_1\cap\cdots\cap O_{m_\alpha}$ for some $O_1,\cdots,O_{m_\alpha}\in\Omega$. Now, if $\mathfrak{T}\in\mathcal{C}$ we know (since $\Omega\subseteq\mathfrak{T}$) that $O_1,\cdots,O_{m_\alpha}\in\mathfrak{T}$ and thus $O_1\cap\cdots\cap O_{m_\alpha}=U_\alpha\in\mathfrak{T}$ for each $\alpha\in\mathcal{A}$. It follows that $U$ is the union of sets in $\mathfrak{T}$ and thus $U\in\mathfrak{T}$. It follows from previous comment that $\mathfrak{G}\subseteq\mathfrak{J}$. The conclusion follows. $\blacksquare$ The actual problem follows immediately from this. c) So, let $g:Y\to A$ be some mapping and suppose that $f_\alpha\circ g:Y\to X_\alpha$ is continuous for each $\alpha\in\mathcal{A}$. Then, given a subbasic open set $U$ in $A$ we have that $U=f_{\alpha_1}^{-1}\left(U_{\alpha_1}\right)\cap\cdots\cap f_{\alpha_n}^{-1}\left(U_{\alpha_n}\right)$ for some $\alpha_1,\cdots,\alpha_n$ and for some open sets $U_{\alpha_1},\cdots,U_{\alpha_n}$ in $X_{\alpha_1},\cdots,X_{\alpha_n}$ respectively. Thus $g^{-1}(U)$ may be written as $\displaystyle g^{-1}\left(\bigcup_{j=1}^{n}f_{\alpha_j}^{-1}\left(U_{\alpha_j}\right)\right)=\bigcup_{j=1}^{n}g^{-1}\left(f_{\alpha_j}^{-1}\left(U_{\alpha_j}\right)\right)=\bigcup_{j=1}^{n}\left(f_{\alpha_j}\circ g\right)^{-1}\left(U_{\alpha_j}\right)$ but since each $f_{\alpha_j}\circ g:Y\to X_{\alpha_j}$ we see that $g^{-1}\left(U\right)$ is the finite union of open sets in $Y$ and thus open in $Y$. It follows that $g$ is continuous. Conversely, suppose that $g$ is continuous then $f_\alpha\circ g:Y\to X_{\alpha}$ is continuous since it’s the composition of continuous maps. d) First note that $\displaystyle f^{-1}\left(\prod_{\alpha\in\mathcal{A}}U_\alpha\right)=\bigcap_{\alpha\in\mathcal{A}}f\alpha^{-1}\left(U_\alpha\right)$ from where it follows that the initial topology under the class of maps $\{f_\alpha\}$ on $A$ is the same as the initial topology given by the single map $f$. So, in general we note that if $X$ is given the initial topology determined by $f:X\to Y$ then given an open set $f^{-1}(U)$ in $X$ we have that $f\left(f^{-1}(U)\right)=U\cap f(X)$ which is open in the subspace $f(X)$. Advertisements June 9, 2010 ## Munkres Chapter 2 Section 19 (Part I) 1. Problem: Suppose that for each $\alpha\in\mathcal{A}$ the topology on $X_\alpha$ is given by a basis $\mathfrak{B}_\alpha$. The collection of all sets of the form $\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha$ Such that $B_\alpha\in\mathfrak{B}_\alpha$ is a basis for $\displaystyle X=\prod_{\alpha\in\mathcal{A}}X_\alpha$ with the box topology, denote this collection by $\Omega_B$. Also, the collection $\Omega_P$ of all sets of the form $\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha$ Where $B_\alpha\in\mathfrak{B}_\alpha$ for finitely many $\alpha$ and $B_\alpha=X_\alpha$ otherwise is a basis for the product topology on $X$. Proof: To prove the first part we let $U\subseteq X$ be open. Then, by construction of the box topology for each $(x_\alpha)_{\alpha\in\mathcal{A}}\overset{\text{def.}}{=}(x_\alpha)\in U$ we may find some $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha$ such that $U_\alpha$ is open in $X_\alpha$ and $\displaystyle (x_\alpha)\in \prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U$. So, then for each $x_\alpha$ we may find some $B_\alpha\in\mathfrak{B}_\alpha$ such that $x_\alpha\in B_\alpha\subseteq U_\alpha$ and thus $\displaystyle (x_\alpha)\in\prod_{\alpha\in\mathcal{A}}B_\alpha\subseteq\prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U$ Noticing that $\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha\in\Omega_B$ and every element of $\Omega_B$ is open finishes the argument. Next, we let $U\subseteq X$ be open with respect to the product topology. Once again for each $(x_\alpha)\in U$ we may find some $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha$ such that $U_\alpha$ is open in $X_\alpha$ for each $\alpha\in\mathcal{A}$ and $U_\alpha=X_\alpha$ for all but finitely many $\alpha$, call them $\alpha_1,\cdots,\alpha_m$. So, for each $\alpha_k,\text{ }k=1,\cdots,m$ we may find some $B_k\in\mathfrak{B}_k$ such that $x\in B_k\subseteq U_k$ and so $\displaystyle (x_\alpha)\in\prod_{\alpha\in\mathcal{A}}V_\alpha\subseteq\prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U$ Where $\displaystyle V_\alpha=\begin{cases}B_k\quad\text{if}\quad \alpha=\alpha_k ,\text{ }k=1,\cdots,m\\ X_\alpha\quad\text{if}\quad x\ne \alpha_1,\cdots,\alpha_m\end{cases}$ Noting that $\displaystyle \prod_{\alpha\in\mathcal{A}}V_\alpha\in\Omega_P$ and $\Omega_P$ is a collection of open subsets of $X$ finishes the argument. $\blacksquare$ 2. Problem: Let $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a collection of topological spaces such that $U_\alpha$ is a subspace of $X_\alpha$ for each $\alpha\in\mathcal{A}$. Then, $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha=Y$ is a subspace of $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=X$ if both are given the product or box topology. Proof: Let $\mathfrak{J}_S,\mathfrak{J}_P$ denote the topologies $Y$ inherits as a subspace of $X$ and as a product space respectively. Note that $\mathfrak{J}_S,\mathfrak{J}_P$ are generated by the bases $\mathfrak{B}_S=\left\{B\cap Y:B\in\mathfrak{B}\right\}$ (where $\mathfrak{B}$ is the basis on $X$ with the product topology), and $\displaystyle \mathfrak{B}_P=\left\{\prod_{\alpha\in\mathcal{A}}O_\alpha:O_\alpha\text{ is open in }U_\alpha,\text{ and equals }U_\alpha\text{ for all but finitely many }\alpha\right\}$ So, let $(x_\alpha)\in B$ where $B\in\mathfrak{B}_S$ then $\displaystyle B=Y\cap\prod_{\alpha\in\mathcal{A}}V_\alpha=\prod_{\alpha\in\mathcal{A}}\left(V_\alpha\cap U_\alpha\right)$ Where $V_\alpha$ is open in $X_\alpha$, and thus $V_\alpha\cap Y$ is open in $U_\alpha$. Also, since $V_\alpha=X_\alpha$ for all but finitely many $\alpha$ it follows that $V_\alpha\cap U_\alpha=U_\alpha$ for all but finitely many $\alpha$. And so $B\in\mathfrak{B}_P$. Similarly, if $B\in\mathfrak{B}_P$ then $\displaystyle B=\prod_{\alpha\in\mathcal{A}}O_\alpha$ Where $O_\alpha$ is open in $U_\alpha$, but this means that $O_\alpha=V_\alpha\cap U_\alpha$ for some open set $V_\alpha$ in $X_\alpha$ and so $\displaystyle B=\prod_{\alpha\in\mathcal{A}}O_\alpha=\prod_{\alpha\in\mathcal{A}}\left(V_\alpha\cap U_\alpha\right)=\prod_{\alpha\in\mathcal{A}}V_\alpha\cap Y\in\mathfrak{B}_S$ From where it follows that $\mathfrak{B}_S,\mathfrak{B}_P$ and thus $\mathfrak{J}_S,\mathfrak{J}_P$ are equal. The case for the box topology is completely analgous. $\blacksquare$ 3. Problem: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a collection of Hausdorff spaces, then $\displaystyle X=\prod_{\alpha\in\mathcal{A}}X_\alpha$ is Hausdorff with either the box or product topologies Proof: It suffices to prove this for the product topology since the box topology is finer. So, let $(x_\alpha),(y_\alpha)\in X$ be distinct. Then, $x_\beta\ne y_\beta$ for some $\beta\in\mathcal{A}$. Now, since $X_\beta$ is Hausdorff there exists disjoint open neighborhoods $U,V$ of $x_\beta,y_\beta$ respectively. So, $\pi_\beta^{-1}(U),\pi_\beta^{-1}(V)$ are disjoint open neighborhoods of $(x_\alpha),(y_\alpha)$ respectively. The conclusion follows. $\blacksquare$ 4. Problem: Prove that $\left(X_1\times\cdots\times X_{n-1}\right)\times X_n\overset{\text{def.}}{=}X\approx X_1\times\cdots\times X_n\overset{\text{def.}}{=}Y$. Proof: Define $\displaystyle \varphi:X\to Y:((x_1,\cdots,x_{n-1}),x_n)\mapsto (x_1,\cdots,x_n)$ Clearly this is continuous since $\pi_{\beta}\circ\varphi=\pi_\beta$ 5. Problem: One of the implications holds for theorem 19.6 even in the box topology, which is it? Proof: If $\displaystyle f:A\to\prod_{\alpha\in\mathcal{A}}X_\alpha$ where the latter is given the box topology then we have that each $\pi_\alpha$ is continuous and thus so is each $\pi_\alpha\circ f:A\to X_\alpha$. $\blacksquare$# 6. Problem: Let $\left\{\bold{x}_n\right\}_{n\in\mathbb{N}}$ be a sequence of points in the product space $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=X$. Prove that $\left\{\bold{x}_n\right\}_{n\in\mathbb{N}}$ converges to $\bold{x}$ if and only if the sequences $\left\{\pi_\alpha(\bold{x}_n)\right\}_{n\in\mathbb{N}}$ coverge to $\pi_\alpha(\bold{x})$ for each $\alpha\in\mathcal{A}$. Is this fact true if one uses the box topology? Proof: Suppose that $U$ is a neighborhood of $\pi_{\alpha}(\bold{x})$ such that $\left(X_\alpha-U\right)\cap\left\{\bold{x}_n:n\in\mathbb{N}\right\}=K$ is infinite. Notice then that if $\pi_{\alpha}(\bold{x}_n)\in K$ that $\bold{x}_n\notin\pi_{\alpha}^{-1}\left(U\right)$ from where it follows that $\pi_{\alpha}^{-1}\left(U\right)$ is a neighborhood of $\bold{x}$ which does not contain all but finitely many values of $\left\{\bold{x}_n:n\in\mathbb{N}\right\}$ contradicting the fact that $\bold{x}_n\to\bold{x}$ in $X$. Conversely, suppose that $\pi_{\alpha}(\bold{x}_n)\to\pi_{\alpha}(\bold{x})$ for each $\alpha\in\mathcal{A}$ and let $\displaystyle \prod_{\alpha\in\mathcal{A}}U_{\alpha}$ be a basic open neighborhood of $\bold{x}$. Then, letting $\alpha_1,\cdots,\alpha_m$ be the finitely many indices such that $U_{\alpha_k}\ne X_k$. Since each $\pi_{\alpha_k}(\bold{x}_n)\to\pi_{\alpha_k}(\bold{x})$ there exists some $n_\ell\in\mathbb{N}$ such that $n_\ell\leqslant n\implies \pi_{\alpha_k}(\bold{x}_k)\in U_k$. So, let $N=\max\{n_1,\cdots,n_k\}$. Now, note that if $\displaystyle \bold{x}_n\notin\prod_{\alpha\in\mathcal{A}}U_\alpha$ then $\pi_{\alpha}(\bold{x}_n)\notin U_\alpha$ for some $\alpha\in\mathcal{A}$. But, since clearly $\pi_{\alpha}(\bold{x}_n)\in X_\alpha$ we must have that $\pi_{\alpha}(\bold{x}_n)\notin U_{\alpha_1}\cup\cdots\cup U_{\alpha_m}$ and thus $n\leqslant N$. It follows that for every $N\leqslant n$ we have that $\displaystyle \bold{x}_n\in\prod_{\alpha\in\mathcal{A}}U_\alpha$. Then, since every neighborhood of $\bold{x}$ contains a basic open neighborhood and the above basic open neighborhood was arbitrary the conclusion follows. Clearly the first part holds in the box topology since there was no explicit mention of the product topology or it’s defining characteristics. That said, the second does not hold. Consider $\displaystyle \bold{x}_n=\prod_{m\in\mathbb{N}}\left\{\frac{1}{n}\right\}$. Clearly each coordinate converges to zero, but $\displaystyle \prod_{m\in\mathbb{N}}\left(\frac{-1}{m},\frac{1}{m}\right)=U$ is a neighborhood of $\bold{0}$ in the product topology. But, if one claimed that for every $n\geqslant N$ (for some $N\in\mathbb{N}$ that $\bold{x}_n\in U$ they’d be wrong. To see this merely note $\displaystyle \frac{1}{N}\notin\left(\frac{-1}{N+1},\frac{1}{N+1}\right)$ and so $\pi_{N+1}\left(\bold{x}_N\right)\notin\pi_{N+1}(U)$ and thus $\bold{x}_{N}\notin U$. 7. Problem: Let $\mathbb{R}^{\infty}$ be the subset of $\mathbb{R}^{\omega}$ consisting of all eventually zero sequences. What is $\overline{\mathbb{R}^{\infty}}$ in the box and product topology? Proof: We claim that in the product topology $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\omega}$. To see this let $\displaystyle \prod_{n\in\mathbb{N}}U_n$ be a basic non-empty open set in $\mathbb{R}^{\omega}$ with the product topology. Since we are working with the product topology we know there are finitely many indices $n_1,\cdots,n_m$ such that $U_{n_k}\ne \mathbb{R}$. So, for each $n_1,\cdots,n_m$ select some $x_{n_k}\in U_{n_k}$ and consider $(x_n)_{n\in\mathbb{N}}$ where $\displaystyle \begin{cases} x_{n_k}\quad\text{if}\quad n=n_1,\cdots,n_m \\ 0 \text{ } \quad\text{if}\quad \text{otherwise}\end{cases}$ Clearly then $\displaystyle (x_n)_{n\in\mathbb{N}}\in \prod_{n\in\mathbb{N}}U_n\cap\mathbb{R}^{\infty}$ and thus every non-empty open set in $\mathbb{R}^{\omega}$ intersects $\mathbb{R}^{\infty}$ and the conclusion follows. Now, we claim that with the box topology that $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}$. To see this let $(x_n)_{n\in\mathbb{N}}\notin\mathbb{R}^{\infty}$. Then, there exists some subsequence $\{x_{\varphi(n)}\}$ of the sequence $\{x_n\}$ which is non-zero. For each $\varphi(n)$ form an interval $I_{\varphi(n)}$ such that $0\notin I_{\varphi(n)}$. Then, consider $\displaystyle \prod_{n\in\mathbb{N}}U_n,\text{ }U_n=\begin{cases} I_{\varphi(m)}\quad\text{if}\quad n=\varphi(m)\\ \mathbb{R}\quad\text{ }\quad\text{if}\quad \text{otherwise}\end{cases}$ Clearly then $\displaystyle \prod_{n\in\mathbb{N}}U_n$ is a neighborhood of $(x_n)_{n\in\mathbb{N}}$ and since each clearly has an infinite subsequence of non-zero values it is disjoint from $\mathbb{R}^{\infty}$. It follows that in $\mathbb{R}^{\omega}$ with the box topology that $\mathbb{R}^{\infty}$ is closed and thus $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}$ as desired. $\blacksquare$ 8. Problem: Given sequences $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ of real numbers with $a_n>0$ define $\varphi:\mathbb{R}^{\omega}\to\mathbb{R}^{\omega}$ by the equation $\varphi:(x_n)_{n\in\mathbb{N}}\mapsto \left(a_n x_n+b_n\right)_{n\in\mathbb{N}}$ Show that if $\mathbb{R}^{\omega}$ is given the product topology that $\varphi$ is a homeomorphism. What happens if $\mathbb{R}^{\omega}$ is given the box topology? Proof: Let us first prove that $\varphi$ is a bijection. To do this we prove something more general… Lemma: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}}$ be two classes of untopologized sets and $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ a collection of bijections $f_\alpha:X_\alpha\to Y_\alpha$. Then, if $\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}$ we have that $\varphi$ is a bijection. Proof: To prove injectivity we note that if $\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\varphi\left((y_\alpha)_{\alpha\in\mathcal{A}}\right)$ Then, $\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}=\left(f_\alpha(y_\alpha)\right)_{\alpha\in\mathcal{A}}$ And by definition of an $\alpha$-tuple this implies that $f_\alpha(x_\alpha)=f_\alpha(y_\alpha)$ for each $\alpha\in\mathcal{A}$. But, since each $f_\alpha:X_\alpha\to Y_\alpha$ is injective it follows that $x_\alpha=y_\alpha$ For each $\alpha\in\mathcal{A}$. Thus, $(x_\alpha)_{\alpha\in\mathcal{A}}=(y_\alpha)_{\alpha\in\mathcal{A}}$ as desired. To prove surjectivity we let $\displaystyle (y_\alpha)_{\alpha\in\mathcal{A}}\in\prod_{\alpha\in\mathcal{A}}Y_\alpha$ be arbitrary. We then note that for each fixed $\alpha\in\mathcal{A}$ we have there is some $x_\alpha\in X_\alpha$ such that $f_\alpha(x_\alpha)=y_\alpha$. So, if $\displaystyle (x_\alpha)_{\alpha\in\mathcal{A}}\in\prod_{\alpha\in\mathcal{A}}X_\alpha$ is the corresponding $\alpha$-tuple of these values we have that $\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}=(y_\alpha)_{\alpha\in\mathcal{A}}$ from where surjectivity follows. Combining these two shows that $\varphi$ is indeed a bijection. $\blacksquare$ Lemma: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}}$ be two classes of non-empty topological spaces and $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ a corresponding class of continuous functions such that $f_\alpha:X_\alpha\to Y_\alpha$. Then, if $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ and $\displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha$ are given the product topologies the mapping $\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}$ is continuous. Proof: Since the codomain is a product space it suffices to show that $\displaystyle \pi_\beta\circ\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to Y_{\beta}$ is continuous for each $\beta\in\mathcal{A}$. We claim though that the diagram  commutes where $\pi^Y_\beta$ and $\pi^X_\beta$ denote the canonical projections from $\displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha$ and $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ to $Y_\beta$ and $X_\beta$ respectively. To see this we merely note that $\displaystyle \pi^Y_\beta\left(\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)\right)=\pi^Y_\beta\left(\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}\right)=f_{\beta}(x_\beta)$ and $f_\beta\left(\pi^X_\beta\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)\right)=f_\beta\left(x_\beta\right)$ which confirms the commutativity of the diagram. But, the conclusion follows since $f_\beta\circ\pi_\beta$ is the composition of two continuous maps (the projection being continuous since $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ is a product space). The lemma follows by previous comment. $\blacksquare$ We come to our last lemma before the actual conclusion of the problem. Lemma: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}}$ be two classes of non-empty topological spaces and $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ a set of homeomorphisms with $f_\alpha:X_\alpha\to Y_\alpha$. Then, $\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}$ is a homeomorphism if $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ and $\displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha$ are given the product topology. Proof: Our last two lemmas show that $\varphi$ is bijective and continuous. To prove that it’s inverse is continuous we note that $\displaystyle \left(\left(\prod_{\alpha\in\mathcal{A}}f_{\alpha}^{-1}\right)\circ\varphi\right)\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\prod_{\alpha\in\mathcal{A}}\left\{f_\alpha^{-1}\left(f_\alpha(x_\alpha)\right)\right\}=(x_\alpha)_{\alpha\in\mathcal{A}}$ And similarly for the other side. Thus, $\displaystyle \varphi^{-1}=\prod_{\alpha\in\mathcal{A}}f_{\alpha}^{-1}$ Which is continuous since each $f_{\alpha}^{-1}:Y_\alpha\to X_\alpha$ is continuous and appealing to our last lemma again. Thus, $\varphi$ is a bi-continuous bijection, or in other words a homeomorphism. The conclusion follows. $\blacksquare$ Thus, getting back to the actual problem we note that if we denote $T_n:\mathbb{R}\to\mathbb{R}:x\mapsto a_nx+b_n$ that each $T_n$ is a homeomorphism. Thus, since it is easy to see that $\displaystyle \varphi=\prod_{n\in\mathbb{N}}T_n$ we may conclude by our last lemma (since we are assuming that we are giving $\mathbb{R}^{\omega}$ in both the domain and codomain the product topology) that $\varphi$ is a homeomorphism. This is also continuous if we give $\mathbb{R}^{\omega}$ the box topology. To see this we merely need to note that $\displaystyle \left(\prod_{\alpha\in\mathcal{A}}f_\alpha\right)^{-1}\left(\prod_{\alpha\in\mathcal{A}}U_\alpha\right)=\prod_{\alpha\in\mathcal{A}}f_\alpha^{-1}\left(U_\alpha\right)$ and thus if all of the $U_\alpha$ are open then so are (since each $f_\alpha$ is continuous) is each $f_\alpha^{-1}(U_\alpha)$ and thus the inverse image of a basic open set is the unrestricted product of open sets and thus basic open. A similar argument works for the inverse function. $\blacksquare$ June 7, 2010 ## Munkres Chapter 2 Section 18 1. Problem: Show that the normal $\varepsilon-\delta$ formulation of continuity is equivalent to the open set version. Proof: Suppose that $\left(\mathcal{M},d\right),\left(\mathcal{N},d'\right)$ are metric spaces and for every $\varepsilon>0$ and every $f(x)\in\mathcal{N}$ there exists some $\delta>0$ such that $f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))$. Then, given an open set $U\subseteq \mathcal{N}$ we have that $f^{-1}(U)$. To see this let $x\in f^{-1}(U)$ then $f(x)\in U$ and since $U$ is open by hypothesis there exists some open ball $B_\varepsilon(f(x))$ such that $B_{\varepsilon}(f(x))\subseteq U$ and thus by assumption of $\varepsilon-\delta$ continuity there is some $\delta>0$ such that $\displaystyle f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))\subseteq U$ and so $B_{\delta}\subseteq f^{-1}(U)$ and thus $x$ is an interior point of $f^{-1}(U)$. Conversely, suppose that the preimage of an open set is always open and let $f(x)\in\mathcal{N}$ and $\varepsilon>0$ be given. Clearly $B_{\varepsilon}(f(x))$ is open and thus $f^{-1}\left(B_{\varepsilon}(f(x))\right)$ is open. So, since $x\in f^{-1}\left(B_{\varepsilon}(f(x))\right)$ there exists some $\delta>0$ such that $B_{\delta}(x)\subseteq f^{-1}\left(B_{\varepsilon}(f(x))\right)$ and so $f\left(B_{\delta}(x)\right)\subseteq f\left(f^{-1}\left(B_{\varepsilon}(f(x))\right)\right)\subseteq B_{\varepsilon}(f(x))$ $\blacksquare$ 2. Problem: Suppose that $f:X\to Y$ is continuous. If $x$ is a limit point of the subset $A$ of $X$, is it necessarily true that $f(x)$ is a limit point of $f(A)$? Proof: No. Consider $(-1,0)\cup(0,1)$ with the suspace topology inherited from $\mathbb{R}$ with the usual topology. Define $f:(-1,0)\cup(0,1)\to D:x\mapsto\begin{cases}0\quad\text{if}\quad x\in(-1,0)\\ 1\quad\text{if}\quad x\in(0,1)\end{cases}$ This is clearly continuous since $f^{-1}(\{1\})=(-1,0)$ and $f^{-1}(\{1\})=(0,1)$ which are obviously open. But, notice that $\frac{-1}{2}$ is a limit point for $(-1,0)$ since given a neighborhood $N$ of $\frac{-1}{2}$ we must have that there is some $(a,b)\cap \left((-1,0)\cup(0,1)\right)\cap \subseteq N$ which contains it. But, $f\left(\frac{-1}{2}\right)=\{0\}$ is not a limit point for $f\left((-1,0)\right)=\{0\}$ since that set has no limit points. $\blacksquare$ 3. Problem: Let $X$ and $X'$ denote a singlet set in the two topologies $\mathfrak{J}$ and $\mathfrak{J}'$ respectively. Let $\text{id}:X'\to X$ be the identity function. Show that a) $\text{id}$ is continuous if and only if $\mathfrak{J}'$ is finer than $\mathfrak{J}$ b) $\text{id}$ is a homeomorphism if and only if $\mathfrak{J}=\mathfrak{J}'$ Proof: a) Assume that $\text{id}$ is continuous then given $U\in\mathfrak{J}$ we have that $\text{id}^{-1}(U)=U\in\mathfrak{J}'$. Conversely, if $\mathfrak{J}'$ is finer than $\mathfrak{J}$ we have that given $U\in\mathfrak{J}$ that $\text{id}^{-1}(U)=U\in\mathfrak{J}'$ b) If $\text{id}$ is a homeomorphism we see that both it and $\text{id}^{-1}=\text{id}:X\to X'$ are continuous and so mimicking the last argument we see that $\mathfrak{J}\subseteq\mathfrak{J}'$ and $\mathfrak{J}'\subseteq\mathfrak{J}$. Conversely, if $\mathfrak{J}=\mathfrak{J}'$ then we now that $U\in\mathfrak{J}\text{ iff }U\in\mathfrak{J}'$ or equivalently that $U\text{ is open in }X\text{ iff }\text{id}(U)=U\text{ is open in }X'$ which defines the homeomorphic property. $\blacksquare$ 4. Problem: Given $x_0\in X$ and $y_0\in Y$ show that the maps $f:X\to X\times Y$ and $g:X\times Y\to Y$ given by $f:x\mapsto (x,y_0)$ and $g:y\mapsto (x_0,y)$ are topological embeddings. Proof: Clearly $f$ and $g$ are continuous since the projection functions are the identity and constant functions. They are clearly injective for if, for example, $f(x)=(x,y_0)=(x',y_0)=f(x')$ then by definition of an ordered pair we must have that $x=x'$. Lastly, the inverse function is continuous since $f^{-1}:X\times \{y_0\}\to X:(x,y_0)\mapsto x$ is the restriction of the projection to $X\times\{y_0\}$. The same is true for $g$. $\blacksquare$ 5. Problem: Show that with the usual subspace topology $[0,1]\approx[a,b]$ and $(0,1)\approx(a,b)$. Proof: Define $f:[0,1]\to[a,b]:x\mapsto (b-a)+a$ and $g:(0,1)\to(a,b):x\mapsto (b-a)+a$. These are easily both proven to be homeomorphisms. $\blacksquare$ 6. Problem: Find a function $f:\mathbb{R}\to\mathbb{R}$ which is continuous at precisely one point. Proof: Define $f:\mathbb{R}\to\mathbb{R}:\begin{cases}x\quad\text{if}\quad x\in\mathbb{Q}\\ 0\quad\text{if}\quad x\notin\mathbb{Q}\end{cases}$ Suppose that $f$ is continuous at $x_0$, then choosing sequences $\{q_n\}_{n\in\mathbb{N}},\{i_n\}_{n\in\mathbb{N}}$ of rational and irrationals numbers respectively both converging to $x_0$. We see by the limit formulation of metric space continuity that $x_0=\lim\text{ }q_n=\lim\text{ }f(q_n)=f(x_0)=\lim\text{ }f(i_n)=\lim\text{ }0=0$ And so if $f$ were to be continuous anywhere it would have to be at $0$. To show that it is in fact continuous at $0$ we let $\varepsilon>0$ be given then choosing $\delta=\varepsilon$ we see that $|x|<\delta\implies |f(x)|\leqslant |x|<\delta=\varepsilon$ from where the conclusion follows since this implies that $\displaystyle \lim_{x\to 0}f(x)=0=f(0)$. $\blacksquare$ 7. Problem: a) Suppose that $f:\mathbb{R}\to\mathbb{R}$ is “continuous from the right”, that is, $\displaystyle \lim_{x\to a^+}f(x)=f(a)$ for each $a\in\mathbb{R}$. Show that $f$ is continuous when considered as a function from $\mathbb{R}_\ell$ to $\mathbb{R}$. b) Can you conjecture what kind of functions $f:\mathbb{R}\to\mathbb{R}$ are continuous when considered as maps as $\mathbb{R}\to\mathbb{R}_\ell$. As maps from $\mathbb{R}_\ell$ to $\mathbb{R}_\ell$? Proof: a) Note that by the assumption that $\displaystyle \lim_{x\to a^+}f(x)=f(a)$ we know that for every $\varepsilon>0$ there exists some $\delta>0$ such that $0\leqslant x-a<\delta$ implies that $|f(x)-f(a)|<\varepsilon$. So, let $U\subseteq\mathbb{R}$ be open and let $a\in f^{-1}(U)$. Then, $f(a)\in U$ and since $U$ is open we see that there is some $\varepsilon>0$ such that $B_{\varepsilon}(f(a))\subseteq U$. But, by assumption there exists some $\delta>0$ such that $0\leqslant x-a<\delta\implies f(x)\in B_{\varepsilon}(f(a))$. But, $\left\{x: 0\leqslant x-a<\delta\right\}=[a,a+\delta)$ and thus $f\left([a,a+\delta)\right)\subseteq B_{\varepsilon}(f(a))\subseteq U$ and thus $[a,a+\delta)\subseteq f^{-1}(U)$ and so $a$ is an interior point for $f^{-1}(U)$ from where it follows that $f^{-1}(U)$ is open and thus $f$ is continuous. b) I’m not too sure, and not too concerned right now. My initial impression is that if $f:\mathbb{R}\to\mathbb{R}_\ell$ is continuous then $f^{-1}([a,b))$ is open which should be hard to do. Etc. 8. Problem: Let $Y$ be an ordered set in the order topology. Let $f,g:X\to Y$ be continuous. a) Show that the set $\Omega=\left\{x\in X:f(x)\leqslant g(x)\right\}$ is closed in $X$ b) Let $h:X\to Y:x\mapsto \max\{f(x),g(x)\}$. Show that $h$ is continuous. Proof: a) Let $x_0\notin\Omega$ then $f(x_0)>g(x_0)$. Suppose first that there is no $g(x_0)<\xi and consider $f^{-1}\left(g(x_0),\infty)\right)\cap g^{-1}\left((-\infty,f(x_0)\right)=U$ This is clearly open in $X$ by the continuity of $f,g$ and $x_0$ is contained in it. Now, to show that $U\cap \Omega=\varnothing$ let $z\in U$ then $f(z)\in f\left(f^{-1}\left(g(x_0),\infty)\right)\cap g^{-1}\left(-\infty,f(x_0)\right)\right)$ which with simplification gives the important part that $f(z)\in (g(x_0),\infty)$ and so $f(z)>g(x_0)$ but since there is no $\xi$ such that $g(x_0)<\xi this implies that $f(z)\geqslant f(x_0)$. Similar analysis shows that $g(z)\in (-\infty,f(x_0))$ and since there is no $\xi$ as was mentioned above this implies that $g(z)\leqslant g(x_0)$. Thus, $g(z)\leqslant g(x_0) and thus $z\notin\Omega$. Now, suppose that there is some $\xi$ such that $g(x_0)<\xi then letting $V=f^{-1}(\xi,\infty)\cap g^{-1}(-\infty,\xi)$ we once again see that $V$ is open and $x_0\in V$. Furthermore, a quick check shows that if $z\in V$ that $f(z)\in(\xi,\infty)$ and so $f(z)>\xi$ and $g(z)\in(-\infty,\xi)$ and so $g(z)<\xi$ and so $f(z)>g(z)$ so that $z\notin\Omega$. The conclusion follows b) Let $\Omega_f=\left\{x\in X:f(x)\geqslant g(x)\right\}$ and $\Omega_g=\left\{x\in X:g(x)\geqslant f(x)\right\}$. As was shown in a) both $\Omega_f,\Omega_g$ are closed and thus define $f\sqcup g:X=\left(\Omega_f\cup\Omega_g\right)\to Y:x\mapsto\begin{cases}f(x)\quad\text{if}\quad x\in\Omega_f\\ g(x)\quad\text{if}\quad x\in\Omega_g\end{cases}$ Notice that since $f,g$ are both assumed continuous and $f\mid_{\Omega_g\cap\Omega_f}=g\mid_{\Omega_f\cap\Omega_g}$ that we may conclude by the gluing lemma that $f\sqcup g$ is in fact continuous. But, it is fairly easy to see that $f\sqcup g=\max\{f(x),g(x)\}$ $\blacksquare$ 9. Problem: Let $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a collection of subset of $X$; let $\displaystyle X=\bigcup_{\alpha\in\mathcal{A}}U_\alpha$. Let $f:X\to Y$ and suppose that $f\mid_{U_\alpha}$ is continuous for each $\alpha\in\mathcal{A}$ a) Show that if the collection $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is finite each set $U_\alpha$ is closed, then $f$ is continuous. b) Find an example where the collection $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is countable and each $U_\alpha$ is closed but $f$ is not continuous. c) An indexed family of sets $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is said to be locally finite if each point of $X$ has a neighborhood that intersects only finitely many elements of $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$. Show that if the family $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is locally finite and each $U_\alpha$ is closed then $f$ is continuous.$

Proof:

a) This follows since if $V\subseteq Y$ is closed then it is relatively easy to check that $\displaystyle f^{-1}(V)=\bigcup_{\alpha\in\mathcal{A}}\left(f\mid_{U_\alpha}\right)^{-1}(V)$ but since each $f\mid_{U_\alpha}$ is continuous we see that $\left(f\mid_{U_\alpha}\right)^{-1}(V)$ is closed in $U_\alpha$. But, since each $U_\alpha$ is closed in $X$ it follows that each $\left(f\mid_{U_\alpha}\right)^{-1}(V)$ is closed in $X$. Thus, $f^{-1}(V)$ is the finite union of closed sets in $X$, and thus closed.

b) Give $[0,1]$ the subspace topology inherited from $\mathbb{R}$ with the usual topology and consider $\left\{f_n\right\}_{n\in\mathbb{N}-\{1,2\}}$ with

$f_n=\iota_{[0,1-\frac{1}{n}]}:\left[0,1-\tfrac{1}{n}\right]\to[0,1]:x\mapsto x$

Clearly each $f_n$ i

Lemma: Let $Y$ be any topological space and $\left\{V_\beta\right\}_{\beta\in\mathcal{B}}$ be a locally finite collection of subsets of $Y$. Then, $\displaystyle \bigcup_{\beta\in\mathcal{B}}\overline{V_\beta}=\overline{\bigcup_{\beta\in\mathcal{B}}V_\beta}$

Proof: The left hand inclusion is standard, so it suffices to prove the right inclusion. So, let $\displaystyle x\in\overline{\bigcup_{\beta\in\mathcal{B}}V_\beta}$ since the collection of sets is locally finite there exists some neighborhood $N$ of $x$ such that it intersects only finitely many, say $V_{\beta_1},\cdots,V_{\beta_n}$, elements of the collection. So, suppose that $x\notin \left(\overline{V_{\beta_1}}\cup\cdots\cup \overline{V_{\beta_n}}\right)$ then $N\cap-\left(\overline{V_{\beta_1}}\cup\cdots\cup\overline{V_{\beta_n}}\right)$ is a neighborhood of $x$ which does not intersect $\displaystyle \bigcup_{\beta\in\mathcal{B}}V_\beta$ contradicting the assumption it is in the closure of that set. $\blacksquare$

Now, once again we let $V\subseteq Y$ be closed and note that $\displaystyle f^{-1}(V)=\bigcup_{\alpha\in\mathcal{A}}\left(f\mid_{U_\alpha}\right)^{-1}(V)$ and each $\left(f\mid_{U_\alpha}\right)^{-1}(V)$ is closed in $U_\alpha$ and since $U_\alpha$ is closed in $X$ we see that $\left(f\mid_{U_\alpha}\right)^{-1}(V)$ is closed in $X$. So, noting that $\left(f\mid_{U_\alpha}\right)^{-1}(V)\subseteq U_\alpha$ it is evident from the assumption that $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is locally finite in $X$ that so is $\left\{\left(f\mid_{U_\alpha}\right)^{-1}(V)\right\}_{\alpha\in\mathcal{A}}$ and thus (for notational convenience) letting $F_\alpha=\left(f\mid_{U_\alpha}\right)^{-1}(V)$ the above lemma implies that

$\displaystyle \overline{f^{-1}(V)}=\overline{\bigcup_{\alpha\in\mathcal{A}}F_\alpha}=\bigcup_{\alpha\in\mathcal{A}}\overline{F_\alpha}=\bigcup_{\alpha\in\mathcal{A}}F_\alpha=f^{-1}(V)$

From where it follows that the preimage of a closed set under $f$ is closed. The conclusion follows. $\blacksquare$

10.

Problem: Let $f:A\to B$ and $g:C\to D$ be continuous functions. Let us define a map $f\times g:A\times C\to B\times D$ by the equation $(f\times g)(a\times c)=f(a)\times g(c)$. Show that $f\times g$ is continuous.

Proof: This follows from noting the two projections of $f\times g$ are $\pi_1\circ(f\times g):A\times B\to C:a\times b\mapsto f(a)$ and $\pi_2\circ(f\times g):A\times B\to D:a\times b\mapsto f(b)$. But, both of these are continuous since $\left(\pi_1(f\times g)\right)^{-1}(U)=f^{-1}(U)\times B$. To see this we note that $x\in f^{-1}(U)\times B$ if and only if $x\in f^{-1}(U)$ which is true if and only if $f(x)=\left(\pi_1\circ(f\times g)\right)(x)\in U$ or in other words $x\in \left(\pi_1\circ(f\times g)\right)^{-1}(U)$. Using this we note that the preimage an open set in $C$ will be the product of open sets by the continuity of $f$. It clearly follows both projections, and thus the function itself are continuous. $\blacksquare$

11.

Problem: Let $F:X\times Y\to Z$. We say that $F$ is continuous in eahc variable separately if for each $y_0\in Y$, the map $h:X\to Z:x\mapsto F(x\times y_0($ is continuous and for each $x_0\in X$ the map $j:Y\to Z:y\mapsto F(x_0\times y)$ is continuous. Show that if $F$ is continuous then $F$ is continuous in each variable separately.

Proof: If $F$ is continuous then clearly it is continuous in each variable since if we denote by $G_{y_0}$ the mapping $G_{y_0}:X\to Z:x\mapsto F(x\times y_0)$ we see that $G_{y_0}=H_{y_0}\circ(F\mid_{X\times\{y_0\}})$ where $H_{y_0}:X\to X\times Y:x\mapsto x\times y_0$ but the RHS is the composition of continuous maps and thus continuous. A similar analysis holds for the other variable.

12.

Problem: Let $F:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ be given by

$\displaystyle F(x\times y)=\begin{cases} \frac{xy}{x^2+y^2}&\mbox{if}\quad x\times y\ne 0\times 0\\ 0 &\mbox{if} \quad x\times y=0\times0\end{cases}$

a) Show that $F$ is continuous in each variable separately.

b) Compute $g:\mathbb{R}\to\mathbb{R}:x\mapsto F(x\times x)$.

c) Show that $F$ is not continuous

Proof:

a) Clearly both $F(x\times y_0)$ and $F(x_0\times y)$ are continuous for $x,y\ne 0$ since they are the quotient of polynomials and the denominator is non-zero. Lastly, they are both continuous at $x,y=0$ since it is trivial to check that $$\displaystyle 0=F(0\times y_0)=F(x_0\times 0)=\lim_{x\to 0}F(x\times y_0)=\lim_{y\to 0}F(x_0\times y)$ b) Evidently $\displaystyle g(x)=F(x\times x)=\begin{cases}\frac{1}{2}\quad\text{if}\quad x\times x\ne 0\\ 0\quad\text{if}\quad x\times x=0\end{cases}$ c) This clearly proves that $F(x\times y)$ is not continuous with $\mathbb{R}^2$ is not continuous since if $\Delta$ is the diangonal we have that $\displaystyle \lim_{(x,y)\in\Delta\to (0,0)}F(x\times y)=\frac{1}{2}\ne F(0\times 0)$ and so in particular $\displaystyle \lim _{(x,y)\to(0,0)}F(x\times y)\ne F(0\times 0)$ 13. Problem: Let $A\subseteq X$; let $f:A\to X$ be continuous and let $Y$ be Hausdorff. Prove that if $f$ may be extended to a continuous function $\overset{\sim}{f}:\overline{A}\to Y$, then $\overset{\sim}{f}$ is uniquely determined by $f$. Proof: I am not too sure what this question is asking, but assuming it’s asking that if this extension existed it’s unique we can do this two ways Way 1(fun way!): Lemma: Let $X$ be any topological space and $Y$ a Hausdorff space. Suppose that $\varphi,\psi:X\to Y$ are continuous and define $A(\varphi,\psi)=\left\{x\in X:\varphi(x)=\psi(x)\right\}$. Then, $A(\varphi,\psi)$ is closed in $X$ Proof: Note that $\varphi\oplus\psi:X\to Y\times Y:x\mapsto (\varphi(x),\psi(x))$ is clearly continuous since $\pi_1\circ(\varphi\oplus\psi)=\varphi$ and $\pi_2\circ(\varphi\oplus\psi)=\psi$. It is trivial then to check that $\displaystyle A(\varphi,\psi)=\left(\varphi\oplus\psi\right)^{-1}(\Delta_Y)$ and since $Y$ is Hausdorff we have that $\Delta_Y\subseteq Y\times Y$ is closed and the conclusion follows. $\blacksquare$ From this we note that if $\varphi,\psi$ agree on $D\subseteq X$ such that $\overline{D}=X$ we have that $X\supseteq A(\varphi,\psi)=\overline{A(\varphi,\psi)}\supseteq\overline{D}=X$ From where it follows that $A(\varphi,\psi)=X$ and so $\varphi=\psi$. So, thinking of $\overline{A}$ as a subspace of $X$ we see that $\text{cl}_{\overline{A}}\text{ }A=Y\cap\text{cl}_{X}\text{ }A=\overline{A}$ and thus clearly $A$ is dense in $\overline{A}$. So, the conclusion readily follows by noting that if $\overset{\sim}{f_1},\overset{\sim}{f_2}$ are two continuous extensions then by definition $A\left(\overset{\sim}{f_1},\overset{\sim}{f_2}\right)\supseteq A$. Way 2(unfun way): Let $\overset{\sim}{f_1},\overset{\sim}{f_2}$ be two extensions of $f$ and suppose there is some $x\in\overline{A}-A(\varphi,\psi)$. Clearly $x\notin A$ and thus $x$ is a limit point of $A$. So, by assumption $\overset{\sim}{f_1}(x)\ne\overset{\sim}{f_2}(x)$ and so using the Hausdorffness of $Y$ we may find disjoint neighborhoods $U,V$ of them respectively. Thus, $\overset{\sim}{f_1}^{-1}(U),\overset{\sim}{f_2}^{-1}(V)$ are neighborhoods of $x$ in $X$. Thus, $\overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V)$ is a neighborhood of $x$. But, clearly there can be no $y\in A\cap\left(\overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V)\right)$ otherwise $\overset{\sim}{f_1}(y)=\overset{\sim}{f_2}(y)\in U\cap V$. It follows that $\overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V)$ is a neighborhood of $x$ disjoint from $A$ which contradicts the density of $A$ in $\overline{A}$. The conclusion follow. $\blacksquare$ May 28, 2010 ## Munkres Chapter 2 Section 17 Theorem 17.11 Problem: Prove that every simply ordered set $X$ with the order topology is Hausdorff. The product of two Hausdorff spaces is Hausdorff. A subspace of a Hausdorff space is Hausdorff. Proof: Let $X$ have the order topology and let $a,b\in X$ be distinct and assume WLOG that $a. If there does not exists a $c\in X$ such that $a then $(-\infty,b),(a,\infty)$ are disjoint neighborhoods of $a,b$ respectively. They are disjoint for to suppose that $x\in(-\infty,b)\cap(a,\infty)$ would imply that $a contradictory to our assumption. If there is some $a then $(-\infty,c),(c,\infty)$ are disjoint neighborhoods of $a,b$ respectively. Let $X,Y$ be Hausdorff and $(x,y),(x',y')\in X\times Y$ be distinct. Since $(x,y)\ne(x',y')$ are distinct it follows we may assume WLOG that $x\ne x'$. But, since $X$ is Hausdorff there exists disjoint neighborhoods $U,V$ of $x,y$. So, consider $U\times Y,V\times Y$ these are clearly neighborhoods of $(x,y),(x',y')$ in $X\times Y$ and to assume $(u,v)\in U\times Y\cap V\times Y$ would imply $\pi_1((u,v))-u\in\pi_1(U\times Y\cap V\times Y)\subseteq U\cap V$. Lastly, suppose that $X$ is Hausdorff and let $Y$ be a subspace of $X$. If $x,y\in Y$ are distinct there exists disjoint neighborhoods $U,V$ of them in $X$. Thus, $U\cap Y,V\cap Y$ are disjoint neighborhoods of them in $Y$. $\blacksquare$ 1. Problem: Let $\mathcal{C}$ be a collection of subsets of the set $X$. Suppose that $\varnothing,X\in\mathcal{C}$, and that the finite union and arbitrary intersection of elements of $\mathcal{C}$ are in $\mathcal{C}$. Prove that the collection $\mathfrak{J}=\left\{X-C:C\in\mathcal{C}\right\}$ is a topology on $X$. Proof: Clearly $X=X-\varnothing$ and $\varnothing=X-X$ are in $\mathfrak{J}$. Now, suppose that $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\subseteq\mathfrak{J}$ then $\displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcap_{\alpha\in\mathcal{A}}\left(X-U_\alpha\right)$ and since $\left\{X-U_\alpha\right\}_{\alpha\in\mathcal{A}}\subseteq\mathcal{C}$ and $\mathcal{C}$ is closed under arbitrary intersection it follows that $\displaystyle \bigcap_{\alpha\in\mathcal{A}}\left(X-U_\alpha\right)=X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha\in\mathcal{C}$ and thus $\displaystyle \bigcup_{\alpha\in\mathcal{A}}U_\alpha\in\mathcal{C}$. Lastly, suppose that $\left\{U_1,\cdots,U_n\right\}\subseteq\mathcal{C}$ then $\displaystyle X-\left(U_1\cap\cdots\cap U_n\right)=\left(X-U_1\right)\cup\cdots\cup\left(X-U_n\right)$ and since $\left\{X-U_1,\cdots,X-U_n\right\}\subseteq\mathcal{C}$ and $\mathcal{C}$ is closed under finite union it follows that $X-\left(U_1\cap\cdots\cap U_n\right)\in\mathcal{C}$ and thus $U_1\cap\cdots\cap U_n\in\mathfrak{J}$. $\blacksquare$ 2. Problem: Show that if $A$ is a closed in $Y$ and $Y$ is closed in $X$ that $A$ is closed in $X$. Proof: Since $A$ is closed in $Y$ and $Y$ is a subspace of $X$ we have that $A=Y\cap G$ for some closed set $G\subseteq X$ but since $Y$ is closed it follows that $A$ is the intersection of two closed sets in $X$ and thus closed in $X$. $\blacksquare$ 3. Problem: Show that if $A$ is closed in $X$ and $B$ is closed in $Y$ then $A\times B$ is closed in $X\times Y$ Proof: This follows immediately from question 9. $\blacksquare$ 4. Problem: Show that if $U$ is open in $X$ and $A$ is closed in $X$, then $U-A$ is open in $X$ and $A-U$ is closed in $X$. Proof: This follows immediately from the fact that $U-A=U\cap\left(X-A\right)$ and $A-U=A\cap\left(X-U\right)$. $\blacksquare$ 5. Problem: Let $X$ be an ordered set in the order topology. Show that $\overline{(a,b)}\subseteq[a,b]$. Under what conditions does equality hold? Proof: Let $x\in(-\infty,a)\cup(b,\infty)$ then $(-\infty,a)\cup(b,\infty)$ is a neighborhood of $x$ disjoint from $(a,b)$ and thus $x\notin\overline{(a,b)}$. Equality will hold when $a,b$ are limit points of $(a,b)$ or said otherwise whenever $a we have that there are some$lated d,e$such that $a. $\blacksquare$ 6. Problem: Let $A,B$ and $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ denote subsets of $X$. Prove that if $A\subseteq B$ then $\overline{A}\subseteq\overline{B}$, $\overline{A\cup B}=\overline{A}\cup\overline{B}$, and $\displaystyle \overline{\bigcup_{\alpha\in\mathcal{A}}U_\alpha}\supseteq\bigcup_{\alpha\in\mathcal{A}}\overline{U_\alpha}$. Give an example where this last inclusion is strict. Proof: We choose to prove the second part first. Let $x\notin\overline{A\cup B}$ then there is a neighborhood $N$ of $x$ such that $N\cap (A\cup B)=(N\cap A)\cup (N\cap B)=\varnothing$ and thus $x\notin \overline{A}\text{ and }x\notin\overline{B}$ and so $x\notin\overline{A}\cup\overline{B}$. Conversely, suppose that $x\notin\overline{A}\cup\overline{B}$ then $x\notin \overline{A}$ and $x\notin\overline{B}$ and so there are neighborhoods $N,G$ of $x$ such that $N\cap A=\varnothing,G\cap B=\varnothing$ clearly then $N\cap G$ is a neighborhood of $x$ such that $(N\cap G)\cap(A\cup B)=\varnothing$ and thus $x\notin\overline{A\cup B}$. Using this, if $A\subseteq B$ then we have that $\overline{B}=\overline{B\cup A}=\overline{B}\cup\overline{A}$ from where it follows that $\overline{A}\subseteq\overline{B}$. Let $\displaystyle x\notin\overline{\bigcup_{\alpha\in\mathcal{A}}U_\alpha}$ then there is a neighborhood $N$ of $x$ such that $\displaystyle N\cap\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcup_{\alpha\in\mathcal{A}}\left(N\cap U_\alpha\right)=\varnothing$ and thus $N\cap U_\alpha=\varnothing$ for every $\alpha\in\mathcal{A}$ and thus $x\notin\overline{U_\alpha}$ for every $\alpha\in\mathcal{A}$ and so finally we may conclude that $\displaystyle x\notin\bigcup_{\alpha\in\mathcal{A}}\overline{U_\alpha}$. To see when inclusion can be strict consider $\mathbb{R}$ with the usual topology. Then, $\displaystyle \mathbb{R}=\overline{\bigcup{q\in\mathbb{Q}}\{q\}}\supsetneq\bigcup_{q\in\mathbb{Q}}\overline{\{q\}}=\bigcup_{q\in\mathbb{Q}}\{q\}=\mathbb{Q}$ $\blacksquare$ 7. Problem: Criticize proof (see book). Proof: There is no guarantee that the $A_\alpha$ for which $U$ intersected will b e the same $A_\alpha$ that $V$ will intersect if you pick another $V$. $\blacksquare$ 8. Problem: Let $A,B$ and $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ denote subsets of $X$. Determine whether the following equations hold, if any equality fails determine which inclusion holds. a) $\overline{A\cap B}=\overline{A}\cap\overline{B}$ b) $\displaystyle \overline{\bigcap_{\alpha\in\mathcal{A}}U_\alpha}=\bigcap_{\alpha\in\mathcal{A}}\overline{U_\alpha}$ c) $\overline{A-B}=\overline{A}-\overline{B}$ Proof: a) Equation does not hold. Consider that $\overline{(-1,0)\cap(0,1)}=\overline{\varnothing}=\varnothing$ but $\overline{(-1,0)}\cap\overline{(0,1)}=[-1,0]\cap[0,1]=\{0\}$. The $\subseteq$ inclusion always holds. b) This follows from a) that equality needn’t hold. Once again the $\subseteq$ inclusion is true. c) This need be true either $\overline{\mathbb{R}-\mathbb{Q}}=\mathbb{R}\ne\overline{\mathbb{R}}-\overline{\mathbb{Q}}=\varnothing$. The $\subseteq$ inclusion holds. $\blacksquare$ 9. Problem: Prove that if $A\subseteq X,B\subseteq Y$ then $\overline{A\times B}=\overline{A}\times\overline{B}$ in the product topology on $X\times Y$. Proof: Let $\displaystyle (x,y)\in\overline{A\times B}$. Let $U\times V$ be a basic open set in $X\times Y$ which contains $(x,y)$. Since $x\in \overline{A},y\in \overline{B}$ we can choose some point $x'\in U\cap A,y'\in V\cap B$. Then, $(x',y')\in U\times V\cap A\times B$. It follows that $\displaystyle (x,y)\in\overline{A\times B}$ Conversely, let $(x,y)\in\overline{A\times B}$. Let $U\subseteq X$ be any set such that $x\in U$. Since $\pi_1^{-1}(U)$ is open in $X\times Y$ it contains some point $(x',y')\in A\times B$. Then, $x'\in U\cap A$. It follows that $x\in\overline{A}$. A similar technique works for $Y$. $\blacksquare$ 10., 11,. 12 Covered in theorem stated and proved at the beginning of the post. 13. Problem: Prove that $X$ is Hausdorff if and only if the diagonal $\Delta=\left\{(x,x):x\in X\right\}$ is closed in $X\times X$ with the product topology. Proof: Suppose that $X$ is Hausdorff then given $x\ne y$ we may find disjoint neighborhoods $U,V$ of them. So, $(x,y)\in U\times V$ and $U\times V\cap \Delta=\varnothing$ since $U\cap V=\varnothing$. It follows that $\Delta$ is closed. Conversely, suppose $\Delta$ is closed in $X\times X$ and $x,y\in X$ are distinct. Then, $(x,y)\notin\Delta$ and so there exists a basic open neighborhood $U\times V$ of $(x,y)$ such that $U\times V\cap \Delta=\varnothing$ and so $U,V$ are neighborhoods of $x,y$ in $X$ which are disjoint. For, to suppose they were not disjoint would to assume that $z\in U\cap V\implies (z,z)\in U\times V$ contradicting the assumption that $U\times V\cap\Delta=\varnothing$. $\blacksquare$ 14. Problem: In the cofinite topology on $\mathbb{R}$ to what point or points does the sequence $\left\{\frac{1}{n}\right\}_{n\in\mathbb{N}}$ converge to? Proof: It converges to every point of $\mathbb{R}$. To see this let $x\in\mathbb{R}$ be arbitrary and let $U$ be any neighborhood of it. Then, $\mathbb{R}-U$ is finite and in particular $\left(\mathbb{R}-U\right)\cap K$ is finite. If it’s empty we’re done, so assume not and let $\frac{1}{n_0}=\min\left(\left(\mathbb{R}-U\right)\cap K\right)$ then clearly for all $n\in\mathbb{N}$ such that $n>n_0$ we have that $\frac{1}{n}<\frac{1}{n_0}$ and thus $\frac{1}{n}$ is in $U$. The conclusion follows. $\blacksquare$ 15. Problem: Prove that the $T_1$ axiom is equivalent to the condition that for each pair of points $x,y\in X$ there are neighborhoods of each which doesn’t contain the other. Proof: Suppose that $X$ is $T_1$ then given distinct $x,y\in X$ the sets $X-\{y\},X-\{x\}$ are obviously neighborhoods of $x,y$ respectively which don’t contain the other. Conversely, suppose the opposite is true and let $y\in X-\{x\}$ then there is a neighborhood $U$ of it such that $x\notin U\implies U\subseteq X-\{x\}$ and thus $X-\{x\}$ is open and $\{x\}$ is therefore closed. $\blacksquare$ 16. Problem: Consider the five topologies on $\mathbb{R}$ given in exercise 7 of section 13 (my section 2). a) Deterime the closure of $K$ under each of these topologies. b) Which of these topologies are Hausdorff? Which are $T_1$ Proof: a) As a reminder the topologies are $\mathfrak{J}_1=\text{usual topology}$ $\mathfrak{J}_2=\text{topology on }\mathbb{R}_K$ $\mathfrak{J}_3=\text{cofinite topology}$ $\mathfrak{J}_4=\text{upper limit topology}$ $\mathfrak{J}_5=\text{left ray topology}$ For the first one we easily see that $\overline{K}=K\cup\{0\}$. For the second one we can see that $\overline{K}=K$. To see this note that $\displaystyle \bigcup_{a is open by definition and thus $K$ being the complement of it is closed. Thus, $\overline{K}=K$ For the third one $\overline{K}=\mathbb{R}$. To see this note that we in a sense already proved this in 14, but for any $x\in\mathbb{R}$ and any neighborhood $N$ of it we have that $N=\mathbb{R}-\{x_1,\cdots,x_n\}$ and thus if $\displaystyle \frac{1}{n_0}=\min\left(\{x_1,\cdots,x_n\}\cap K\right)$ (assuming it’s non-empty) we see that $n>n_0\implies \frac{1}{n}\notin\{x_1,\cdots,x_n\}\implies \frac{1}{n}\in N$. Thus, given any point of $\mathbb{R}$ and any neighborhood $N$ of it we have that $N\cap K$ is infinite, and thus clearly $x\in\overline{K}$. The conclusion follows from that. For the fourth topology we note that $\displaystyle \mathbb{R}-K=(-\infty,0]\cup\bigcup_{n\in\mathbb{N}}\left(\frac{1}{n+1},\frac{1}{n}\right)\cup(1,\infty)$ which is open in $\mathbb{R}$ with the upper limit topology. Remember that $\displaystyle (a,b)=\bigcup_{a For the last one $\overline{K}=[0,\infty)$. Clearly $K\subseteq[0,\infty)$ and since $[0,\infty)$ is closed and $\overline{K}$ is the intersection of all closed supersets of $K$ it follows that $\overline{K}\subseteq[0,\infty)$. Now, suppose that $x\notin[0,\infty)$ then $x\in(-\infty,0)$ which is a neighborhood of $x$ which doesn’t intersect $K$ and thus $x\notin\overline{K}$. So, $[0,\infty)\subseteq\overline{K}$. The conclusion follows. b) Lemma: If $X$ is Hausdorff, then it is $T_1$ Proof: Let $x\in X$ and $y\in X-\{x\}$ by assumption there exists disjoint neighborhoods $U,V$ of $x,y$ respectively and so clearly $y\in V\subseteq X-\{x\}$ and thus $X-\{x\}$ is open and so $\{x\}$ is closed. $\blacksquare$ $\mathfrak{J}_1$: Clearly the first topology is Hausdorff since it is an order topology and we proved in the initial post that all order topologies are Hausdorff. Lemma: Let $X$ be a set and $\mathfrak{J},\mathfrak{J}'$ two topologies on $X$ such that $\mathfrak{J}'$ is finer than $\mathfrak{J}$. Then, if $X$ is Hausdorff with the $\mathfrak{J}$ topology it is Hausdorff with the $\mathfrak{J}'$ topology. Proof: Clearly if $x,y\in X$ are distinct we may find disjoint neighborhoods $U,V\in\mathfrak{J}$ of them in the topology given by $\mathfrak{J}$ and thus the same neighborhoods work in consideration of the topology given by $\mathfrak{J}'$. $\blacksquare$ $\mathfrak{J}$: From this lemma it follows that $\mathbb{R}_K$ having a finer topology than $\mathbb{R}$ with the usual topology is Hausdorff $\mathfrak{J}_3$: The cofinite topology is $T_1$ but not Hausdorff. To see that it’s $T_1$ it suffices to prove that $\{x\}$ is closed for any $x\in\mathbb{R}$. But, this is trivial since $\mathbb{R}-\{x\}$ being the complement of a finite set is open, thus $\{x\}$ closed. But, it is not Hausdorff since it cannot support two disjoint, non-empty open sets. To see this suppose that $U$ is open in $\mathbb{R}$ with the cofinite topology then $\mathbb{R}-U$ is finite and since a disjoint set $V$ would have to be a subset of $\mathbb{R}-U$ it follows that $V$ is finite and thus it’s complement not finite. Thus, $V$ is not open. $\mathfrak{J}_4$: Once again this topology finer than that of $\mathbb{R}$ with the usual topology since $\displaystyle (a,b)=\bigcup_{c $\mathfrak{J}_5$: This isn’t even $T_1$. To see this we must merely note that if $U$ is any set containing $1$ we must have that there is some basic open set $(-\infty,\alpha),\text{ }\alpha>1$ such that $1\in(-\infty,\alpha)\subseteq U$, but this means that $0\in U$. So, there does not exist a neighborhood of $1$ which does not contain $0$. 17. Problem: Consider the lower limit topology on $\mathbb{R}$ and the topology given by the basis $\mathcal{C}=\left\{[a,b):a. Determine the closure of the intervals $A=(0,\sqrt{2})$ and $B=(\sqrt{2},3)$ in these two topologies. Proof: We first prove more generally that if $(a,b)\subseteq\mathbb{R}_\ell$ then $\overline{(a,b)}=[a,b)$. To see this we first note that $[a,b)$ is open since $\mathbb{R}-[a,b)=(-\infty,a)\cup[b,\infty)$ which we claim is open. To see this we note that $\displaystyle (-\infty,a)\cup[b,\infty)=\bigcup_{xb}[b,y)$. So, since $\overline{(a,b)}$ is the intersection of all closed supersets of $(a,b)$ and $[a,b)\supseteq(a,b)$ is closed we see that $\overline{(a,b)}\subseteq[a,b)$. So, we finish the argument by showing that $\overline{(a,b)}\ne(a,b)$. To see this we show that $a$ is a limit point for $(a,b)$ from where the conclusion will follow. So, let $N$ be any neighborhood of $a$, then we may find some basic open neighborhood $[\alpha,\beta)$ such that $a\in[\alpha,\beta)\subseteq N$, but clearly $[\alpha,\beta)\cap(a,b)$ contains infinitely many points from where it follows that $a$ is a limit point of $(a,b)$. From this we may conclude for $\mathbb{R}_\ell$ that $\overline{(0,\sqrt{2})}=[0,\sqrt{2})$ and $\overline{(\sqrt{2},3)}=[\sqrt{2},3)$ We now claim that in the topology generated by $\mathcal{C}$ that $\overline{(0,\sqrt{2})}=[0,\sqrt{2}]$ and $\overline{(\sqrt{2},3)}=[\sqrt{2},3)$. More generall, let us prove that in this topology $\overline{(a,b)}=\begin{cases}[a,b]\quad\text{if}\quad b\notin\mathbb{Q}\\ [a,b)\quad\text{if}\quad b\in\mathbb{Q}\end{cases}$ Clearly if $\alpha then choosing some rational number $q<\alpha$ and some rational number $\alpha then $\alpha\in[q,p)$ and $[q,p)\cap(a,b)=\varnothing$ so that $\alpha\notin\overline{(a,b)}$. Also, if $\alpha>b$ then choosing some $p\in\mathbb{Q}$ such that $b we see that $\alpha\in[p,\alpha+1)$ and $[p,\alpha+1)\cap(a,b)=\varnothing$. Thus, the only possibilities for $\overline{(a,b)}$ are $(a,b),[a,b),[a,b]$. But, just as before if $N$ is any neighborhood $a$ we may find some open basic neighborhood$latex [p,q)$such that $a\in[p,q)\subseteq N$ but clearly $[p,q)\cap(a,b)\ne\varnothing$ and thus since $N$ was arbitrary it follows that $a\in\overline{(a,b)}$. So, now we split into the two cases. First assume that $b\notin\mathbb{Q}$ then given any neighborhood $N$ of $b$ we may find some basic neighborhood $[p,q)$ such that $b\in[p,q)\subseteq N$, but since $b\notin\mathbb{Q}$ we see that $b\ne p$ from where it follows that $[p,q)\cap(a,b)\ne\varnothing$ and thus since $N$ was arbitrary it follows that $b\in\overline{(a,b)}$ and thus $\overline{(a,b)}=[a,b]$. Now suppose that $b\in\mathbb{Q}$, then $[b,b+1)$ is clearly a neighborhood of $b$ that doesn’t intersect $(a,b)$ and thus $b\notin{(a,b)}$ so that $\overline{(a,b)}=[a,b)$ The conclusion follows. $\blacksquare$ Problem: If $A\subseteq X$, we define the boundary of $A$ by $\partial A=\overline{A}-\overline{X-A}$. a) Prove that $A^\circ$ and $\partial A$ are disjoint and $\overline{A}=A^\circ\cup\partial A$ b) Prove that $\partial A=\varnothing$ i and only if $A$ is both open and closed c) Prove that $U$ is open if and only if $\partial U=\overline{U}-U$ d) If $U$ is open, is it true that $U=\left(\overline{U}\right)^{\circ}$? Proof: a) Clearly if $x\in A^{\circ}$ then there exists a neighborhood of $x$ whose intersection with the complement of $A$ is empty, thus $x\notin\partial A$. Now, let $x\in\overline{A}$ now if there exists a neighborhood $N$ of $x$ such that $N\subseteq A$ then $x\in A^{\circ}$ and if not then every neighborhood of $x$ contains points of $X-A$ and since $x\in D(A)$ it follows that it also contains points of $A$. Thus, either $x\in A^{\circ}$ or $x\in\partial A$, thus $x\in A^{\circ}\cup\partial A$. Conversely, if $x\in A^{\circ}\partial A$ then either there exists a neighborhood $N$ of $x$ such that $N\subseteq A$ and thus $x\in A$. Conversely, if $x\in\partial A$ then for every neighborhood $N$ of $x$ we have that $N\cap A,N\cap (X-A)=\ne\varnothing$ and in particular $N\cap A\ne\varnothing$ and so $x\in\overline{A}$. b) Suppose first that $\partial A=\varnothing$. If $A=\varnothing$ we’re done, so assume not and let $x\in A$. Since $x\notin\partial A$ there is a neighborhood $N$ of $x$ such that $N\cap A\varnothing\text{ or }N\cap (X-A)=\varnothing$ but since $x\in N\cap A$ it follows that $N\subseteq A$ and thus $A$ is open. Conversely, letting $x\in X-A$ we see that $x\notin\partial A$ and so by the same logic there exists a neighborhood $N$ of $x$ such that $N\subseteq X-A$ and thus $X-A$ is open and so $A$ closed. Conversely, suppose that $A$ is both open and closed and suppose that $\partial A\ne\varnothing$. If $x\in\partial A\cap A$ then for every neighborhood $N$ of $x$ we must have that $N\cap (X-A)\ne\varnothing$ and thus $x\notin A^{\circ}=A$, which is a contradiction. Conversely, if $x\in (X-A)\cap\partial A$ then for every neighborhood $N$ of $x$ we must have that $N\cap A\ne\varnothing$ and thus $x\notin (X-A)^{\circ}=X-A$ which is a contradiction. c) Suppose that $U$ is open and let $x\in\partial U$, then every neighborhood $N$of $x$ contains points of $U$ by definition, but since $x\notin U$ (it can’t be an interior point, thus $x\notin U^{\circ}=U$) they must be points of $U-\{x\}$ and thus $x\in D(U)$. So, $x\in \overline{U}-U$. Conversely, if $x\in \overline{U}-U$ then $x$ must be a limit point of $U$ which is not in $U$ and thus every neighborhood $x$ contains points of $U$ and $X-U$ so $x\in\partial U$ d) No, it is not true. With the usual topology on $\mathbb{R}$ the set $(-1,0)\cup(0,1)$ is open, but $\left(\overline{(-1,0)\cup(0,1)}\right)^{\circ}=[-1,1]^\circ=(-1,1)$ We choose to omit 18, 19, 21 on the grounds that they are monotonous and easy, monotonous and easy, and way, way to long. May 22, 2010 ## Munkres Chapter 2 Section 1 This begins a substantial effort to complete all of (except the first chapter) the problems in James Munkre’s Topology I have chosen to start at chapter 2 considering the first chapter is nothing but prerequisites. Thus, without a minute to spare let us being. 1. Problem: Let $X$ be a topological space, let $A$ be a subset of $X$. Suppose that for each $x\in A$ there is an open set $U_x$ containing $x$ such that $U\subseteq A$. Prove that $A$ is open. Proof: We claim that $\displaystyle A=\bigcup_{x\in A}U_x=\overset{\text{def.}}{=}\Omega$ but this is obvious since for each $x\in A$ we have that $x\in U_x\subseteq\Omega$. Conversely, since each $U_x\subseteq A$ we have that the union of all of them is contained in $A$, namely $\Omega\subseteq A$. Thus, $A$ is the union of open sets and thus open. $\blacksquare$ 2. Problem: Compare the nine topologies on $\{a,b,c\}$ given in example 1. Solution: This is simple. 3. Problem: Show that given a set $X$ if we denote $\mathfrak{J}$ to be cocountable topology (a set is open if it’s complement is countable or the full space) that $\left(X,\mathfrak{J}\right)$ is a topological space. Is it still a topological space if we let $\mathfrak{J}=\left\{U\in\mathcal{P}(X):X-U\text{ is infinite, empty, or all of }X\right\}$? Proof: Clearly for the first part $\varnothing,X\in\mathfrak{J}$. Now, if $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is a collection of open sets then we note that $X-U_\alpha$ is finite and thus $\displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcap_{\alpha\in\mathcal{A}}\left(X-U_\alpha\right)\subseteq U_{\alpha_0}$ for any $\alpha_0$. Thus, $\displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha$ is finite and thus in $\mathfrak{J}$. Now, if $U_1,\cdots,U_n\in\mathfrak{J}$ we have that $X-(U_1\cap\cdots\cap U_n)=(X-U_1)\cup\cdots\cup (X-U_n)$ and thus $X-(U_1\cap\cdots\cap U_n)$ is the finite union of finite sets and thus finite, so $U_1\cap\cdots\cap U_n\in\mathfrak{J}$. If we redefine the topology as described it is not necessarily a topology. For example, give $\mathbb{N}$ that topology and note that $\left\{\{n\}\right\}_{n\in\mathbb{N}-\{1\}}$ is a collection of elements of $\mathfrak{J}$ but $\displaystyle \bigcup_{n\in\mathbb{N}-\{1\}}\{n\}=\mathbb{N}-\{1\}$ whose complement is neither infinite, empty, or the full space. Thus, this isn’t a topology. 4. Problem: a) If $\left\{\mathfrak{J}_\alpha\right\}_{\alpha\in\mathcal{A}}$ is a family of topologies on $X$, show that $\displaystyle \bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha$ is a topology on $X$. Is $\displaystyle \bigcup_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha$? b) Let $\left\{\mathfrak{J}\right\}_\alpha$ be a family of topologies on $X$. Show that there is unique topology on $X$ containing all the collections $\mathfrak{J}_\alpha$, and a unique largest topology contained in all of the $\mathfrak{J}_\alpha$. c) If $X=\{a,b,c\}$, let $\mathfrak{J}_1=\{\varnothing,X,\{a\},\{a,b\}\}$ and $\mathfrak{J}_2=\left\{\varnothing,X,\{a\},\{b,c\}\right\}$. Find the smallest topology containing $\mathfrak{J}_1,\mathfrak{J}_2$ and the larges topology contained in $\mathfrak{J}_1,\mathfrak{J}_2$. Proof: a) Let $\displaystyle \Omega\overset{\text{def.}}{=}\bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha$ and let $\left\{U_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\Omega$ be arbitrary. Then, by assumption we have that $\left\{U_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\mathfrak{J}_\alpha$ for every $\alpha\in\mathcal{A}$ but since this was assumed to be a topology we have that for every $\alpha\in\mathcal{A}$ that $\displaystyle \bigcup_{\beta\in\mathcal{B}}U_\beta\subseteq\mathfrak{J}_\alpha$ and thus $\displaystyle \bigcup_{\beta\in\mathcal{B}}U_\beta\in\Omega$. Now, if $\{U_1,\cdots,U_n\}\subseteq\Omega$ we must have that $\{U_1,\cdots,U_n\}\subseteq\mathfrak{J}_\alpha$ for every $\alpha$ and thus $U_1\cap\cdots\cap U_n\in\mathfrak{J}_\alpha$ for every $\alpha\in\mathcal{A}$ and thus $U_1\cap\cdots\cap U_n\in\Omega$. Thus, noting that for every $\alpha\in\mathcal{A}$ we must have that $\varnothing,X\in\mathfrak{J}_\alpha$ the conclusion follows. No, the union of two topologies needn’t be a topology. Let $\mathfrak{J}_1,\mathfrak{J}_2$ be defined as in part c and note that $\mathfrak{J}_1\cup\mathfrak{J}_2=\left\{\varnothing,X,\{a\},\{a,b\},\{b,c\}\right\}$ but $\{a,b\}\cap\{b,c\}=\{c\}\notin\mathfrak{J}_1\cup\mathfrak{J}_2$ b) This follows immediately from part a. For the first part let $\Omega=\left\{\mathfrak{J}\in\text{Top }X:\mathfrak{J}_\alpha\subseteq\mathfrak{J},\text{ for all }\alpha\in\mathcal{A}\right\}$ (where $\text{Top }X$ is the set of all topologies) and let $\mathfrak{T}=\bigcap_{\mathfrak{J}\in\Omega}\mathfrak{J}$, this clearly satisfies the conditions. For the second one merely take $\displaystyle \bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha$ c) We (as pointed out very poignantly in the previous problem) just take the union of the two topologies. So, we claim that the desired smallest topology is in fact $\left\{X,\varnothing,\{a\},\{b\},\{a,b\},\{a,c\}\right\}$. But this is just computation and we leave it to the reader. For the second part just intersect the two topologies. 5. Problem: Show that if $\mathcal{A}$ is a base for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ which contain $\mathcal{A}$. Prove the same if $\mathcal{A}$ is a subbase Proof: Let $\Omega$ be the intersection of all topologies on $X$ which contain $\mathcal{A}$ and $\mathfrak{J}_g$ the topology generated by $\mathcal{A}$. Clearly $\Omega\subseteq\mathfrak{J}_g$ since $\mathfrak{J}_g$ is itself a topology on $X$ containing $\mathcal{A}$. Conversely, let $U\in\mathfrak{J}_g$ we show that $U\in\mathfrak{J}$ where $\mathfrak{J}$ is any topology on $X$ containing $X$. But, this is obvious since $\displaystyle U=\bigcup_{B\in\mathcal{B}}B$ for some $\mathcal{B}\subseteq\mathcal{A}$ and thus $U$ is the union of open sets in $\mathfrak{J}$ and thus in $\mathfrak{J}$. The conclusion follows. Next, let $\Omega,\mathfrak{J}_g$ be above except now $\mathcal{A}$ is a subbase. For the same reasons as above we have that $\Omega\subseteq\mathfrak{J}_g$. Conversely, for any topology $\mathfrak{J}$ containing $\mathcal{A}$ we have that if $U\in\mathfrak{J}_g$ then $\displaystyle U=\bigcup_{\beta\in\mathcal{B}}V_\beta$ where each $V_\beta$ is the finite union of elements of $\mathcal{A}$. But, by construction it follows that each $V_\beta$ is open (it is the finite intersection of open sets in $\mathfrak{J}$) and thus $U$ is the union of open sets in $\mathfrak{J}$ and thus $V\in\mathfrak{J}$. $\blacksquare$ 6. Problem: Show that the topologies on $\mathbb{R}_K$ and the Sorgenfrey line aren’t comparable Proof: See the last part of the next problem 7. Problem: Consider the following topologies on$\mathbb{R}$: $\mathfrak{J}_1=\text{usual topology}$ $\mathfrak{J}_2=\text{topology on }\mathbb{R}_K$ $\mathfrak{J}_3=\text{cofinite topology}$ $\mathfrak{J}_4=\text{topology having the set set of all }(a,b]\text{ as a base}$ $\mathfrak{J}_5=\text{the topology having all sets }(-\infty,a)\text{ as a base}$ For each determine which of the others contain it. Solution: $\mathfrak{J}_1\subseteq\mathfrak{J}_2$:Clearly we have that $\mathfrak{J}_1\subseteq\mathfrak{J}_2$ since the defining open base for $\mathfrak{J}_1$ is contained entirely in $\mathfrak{J}_2$. $\mathfrak{J}_1\supseteq\mathfrak{J}_3$: But, $\mathfrak{J}_1\not\subseteq\mathfrak{J}_3$ since $(0,1)\in\mathfrak{J}_1$ but $\mathbb{R}-(0,1)\simeq\mathbb{R}$ and thus not in $\mathfrak{J}_3$. Now, to prove the inclusion indicated we know that for each open set $U$ in the cofinite topology we have that $U=\mathbb{R}-\{x_1,\cdots,x_n\}$ and so if we assume WLOG that $x_1<\cdots then $\displaystyle U=\bigcup_{ax_n}(x_n,b)$ And thus $U$ is open in the usual topology. $\mathfrak{J}_1\subseteq\mathfrak{J}_4$: But, $\mathfrak{J}_1\subseteq\mathfrak{J}_4$. To see this it suffices to show that $(a,b)$ is open in $\mathfrak{J}_4$ since this is a base for $\mathfrak{J}_1$. But, to see this we must merely note that $\displaystyle (a,b)=\bigcup_{c. $\mathfrak{J}_1\supseteq\mathfrak{J}_5$: Lastly, $\mathfrak{J}_1\supseteq\mathfrak{J}_5$. To see this we must merely note that $\displaystyle (-\infty,b)=\bigcup_{a. For $\mathfrak{J}_2$ the result is obvious except possibly how it relates to $\mathfrak{J}_3$. But, in fact they aren’t comparable. To see this we first show that $(0,1]$ is not open in $\mathbb{R}_K$. To see this we show it can’t be written as the union of sets of the form $(a,b)$ and $(c,d)-K$ but it clearly suffices to do this for the latter sets. Now, to see that $(0,1]$ can’t be written as the union of sets of the form $(a,b)$ we recall from basic real number topology that $(0,1]$ is not open ($1$ is not an interior point) and thus it can’t be written as the union of intervals or it would be open in the usual topology on $\mathbb{R}$. Also, consider $(-1,1)-K$. 8. Problem: Show that the countable collection $\mathfrak{B}=\left\{(a,b):a is a base for the usual topology on $\mathbb{R}$ Proof: This follows from the density of $\mathbb{Q}$. It suffices to show that given any $x\in\mathbb{R}$ and any $(a,b)\supseteq\{x\}$ that there is some element $(p,q)\in\mathfrak{B}$ such that $x\in(p,q)\subseteq(a,b)$. But, from basic analysis we know there is some rational number $q$ such that $a and similarly there is some $p\in\mathbb{Q}$ such that $x. Thus, $x\in(p,q)\subseteq(a,b)$. The conclusion follows. $\blacksquare$. 9. Problem: Show that the collection $\mathcal{C}=\left\{[a,b):a generates a different topology from the one on $\mathbb{R}_\ell$ (the lower limit topology)$.

Proof: Clearly $[\sqrt{2},3)$ is open in $\mathbb{R}_\ell$ but we show that it can’t be written as the union of elements of $\mathcal{C}$. So, suppose that $\displaystyle \bigcup_{\alpha\in\mathcal{A}}[a_\alpha,b_\alpha)=[\sqrt{2},3)$ where $\left\{[a_\alpha,b_\alpha)\right\}_{\alpha\in\mathcal{A}}\subseteq\mathcal{C}$. Then, there exists some $[a_\alpha,b_\alpha)$ such that $\sqrt{2}\in[a_\alpha,b_\alpha)$. Now, since $[a_\alpha,b_\alpha)\subseteq[\sqrt{2},b_\alpha)$ we must have that $a_\alpha\geqslant \sqrt{2}$ but $a_\alpha\neq\sqrt{2}$ and thus $a_\alpha>\sqrt{2}$ and thus $\sqrt{2}\notin[a_\alpha,b_\alpha)$. Contradiction. $\blacksquare$

May 20, 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