Exercise 20.8

Question

Let $X$ be the subset of $\mathbb{R}^\omega$ consisting of all sequences $\mathbf{x}$ such that $\sum x_i^2$ converges. Then the formula
\begin{equation*}
  d(\mathbf{x},\mathbf{y}) = \left[ \sum_{i=1}^\infty (x_i-y_i)^2 \right]^{1/2}
\end{equation*}
defines a metric on $X$. On $X$ we have the three topologies it inherits from the box, uniform, and product topologies on $\mathbb{R}^\omega$. We have also the topology given by the metric $d$, which we call the $\ell^2$-\emph{topology}.
\begin{enumerate}[label=(\alph*)]
\item Show that on $X$, we have the inclusions
  \begin{center}
    box topology $\supset$ $\ell^2$-topology $\supset$ uniform topology.
  \end{center}
\item The set $\mathbb{R}^\infty$ of all sequences that are eventually zero is contained in $X$. Show that the four topologies that $\mathbb{R}^\infty$ inherits as a subspace of $X$ are all distinct.
\item The set
  \begin{equation*}
    H = \prod_{n \in \mathbb{Z}_+} [0,1/n]
  \end{equation*}
  is contained in $X$; it is called the \emph{Hilbert cube}. Compare the four topologies that $H$ inherits as a subspace of $X$.
\end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $(a)$] box topology $\supset$ $\ell^2$-topology. Let $\mathcal{U} i \mathbf{x}$ be a basis element in the $\ell^2$-topology. Then, there exists a basis element $U = B_d(\mathbf{x},\epsilon) \subseteq \mathcal{U}$ of the $\ell^2$-topology. We claim that $V = X \cap \prod (x_i-\epsilon/2^{i/2},x_i+\epsilon/2^{i/2}) i \mathbf{x}$ basis element of the box topology is contained in $U$. Suppose $\mathbf{y} \in V$; then \begin{equation*} d(\mathbf{x},\mathbf{y})^2 = \sum_{i=1}^\infty (x_i-y_i)^2 < \sum_{i=1}^\infty \frac{\epsilon^2}{2^i} = \epsilon^2 \implies \mathbf{y} \in U, \end{equation*} i.e., $V \subseteq U \subseteq \mathcal{U}$, and box topology $\supset$ $\ell^2$-topology by Lemma $13.3$. \par $\ell^2$-topology $\supset$ uniform topology. Let $\mathcal{U} i \mathbf{x}$ be a basis element in the uniform topology. Then, there exists a basis element $U = X \cap B_{\overline{ ho}}(\mathbf{x},\epsilon) \subseteq \mathcal{U}$ of the uniform topology. We claim $V = B_d(\mathbf{x},\epsilon)$ basis element of the $\ell^2$-topology is contained in $U$. If $\epsilon > 1$, then trivially $V \subseteq U = X$, so we assume $\epsilon \le 1$. Suppose $\mathbf{y} \in V$; then \begin{equation*} \overline{ ho}(\mathbf{x},\mathbf{y}) = \sup |x_i-y_i| \le d(\mathbf{x},\mathbf{y}) < \epsilon \implies \mathbf{y} \in U, \end{equation*} i.e., $V \subseteq U \subseteq \mathcal{U}$, and $\ell^2$-topology $\supset$ uniform topology by Lemma $13.3$. \end{proof} \begin{proof}[Proof of $(b)$] By $(a)$ and Theorem $20.4$, we have the inclusions \begin{center} box topology $\supset$ $\ell^2$-topology $\supset$ uniform topology $\supset$ product topology, \end{center} since if $\mathcal{T}_1 \supset \mathcal{T}_2$ in the ambient space, we would also have $\mathcal{T}_1' \supset \mathcal{T}_2'$, where $\mathcal{T}_i'$ are the subspace topologies induced by $\mathcal{T}_i$ on $X$, by the fact that the open sets in the subspace $X$ are the open sets of the ambient space intersected with $X$. We claim these are strict inclusions. \par box topology $\supsetneq$ $\ell^2$-topology. Consider the open set $U = \mathbb{R}^\infty \cap \prod (-1/i,1/i) i 0$ in the box topology. Consider any neighborhood $\mathcal{V} i 0$; then, there exists some $V = \mathbb{R}^\infty \cap B_d(0,\epsilon)$ for $\epsilon > 0$ contained in $\mathcal{V}$. We can find $N \in \mathbb{N}$ such that $1/N < \epsilon$, and so $\mathbf{x}$ such that $x_i = 0$ for all $i$ except $x_N = 1/N$ is contained in $V$ since $d(0,\mathbf{x}) = 1/N < \epsilon$ and therefore $\mathcal{V}$, but not in $U$. Hence, $U$ is open in the box topology but not in the $\ell^2$-topology by p.~78, and so box topology $\supsetneq$ $\ell^2$-topology. \par $\ell^2$-topology $\supsetneq$ uniform topology. Consider the open set $U = \mathbb{R}^\infty \cap B_d(0,1)$ in the $\ell^2$-topology. Consider any neighborhood $\mathcal{V} i 0$; then, there exists some $V = \mathbb{R}^\infty \cap B_{\overline{ ho}}(0,\epsilon)$ for $\epsilon > 0$ contained in $\mathcal{V}$. We can find $N \in \mathbb{N}$ such that $N\epsilon^2 > 4$, and so $\mathbf{x}$ such that $x_i = \epsilon/2$ for all $1 \le i \le N$ and $0$ otherwise is contained in $V$ and therefore $\mathcal{V}$, yet \begin{equation*} d(\mathbf{x},0)^2 = \sum_{i=1}^\infty x_i^2 = N\frac{\epsilon^2}{4} > 1 \implies d(\mathbf{x},0) > 1, \end{equation*} and so $\mathbf{x} otin U$. Hence, $U$ is open in the $\ell^2$-topology but not in the uniform topology by p.~78, and so $\ell^2$-topology $\supsetneq$ uniform topology. \par uniform topology $\supsetneq$ product topology. Consider the open set $U = \mathbb{R}^\infty \cap B_{\overline{ ho}}(0,1)$ in the uniform topology. Consider any $V = \mathbb{R}^\infty \cap \prod U_\alpha i 0$ where $U_\alpha = \mathbb{R}$ for all but finitely many $\alpha$. Let $N$ be such that $U_N = \mathbb{R}$; then, $\mathbf{x}$ such that $x_i = 0$ for all $i$ except $|x_N| \ge 1$ is in $V$ but not in $U$. Hence, $U$ is open in the uniform topology but not in the product topology by p.~78, and so uniform topology $\supsetneq$ product topology. \end{proof} \begin{proof}[Solution for $(c)$] We claim that \begin{center} box topology $\supsetneq$ ($\ell^2$-topology $=$ uniform topology $=$ product topology), \end{center} i.e., the box topology is strictly finer than the other topologies, which are equal. \par To show the equality, it suffices to show product topology $\supset$ $\ell^2$-topology, for then we have $\ell^2$-topology $\supset$ uniform topology $\supset$ product topology $\supset$ $\ell^2$-topology by the same argument as in $(b)$, and so we must have equality throughout. So, consider $\mathcal{U} i \mathbf{x}$ open in the $\ell^2$-topology; then, we can find a basis element $U = H \cap B_d(\mathbf{x},\epsilon) \subseteq \mathcal{U}$ of the $\ell^2$-topology for some $\epsilon > 0$. Let $\delta = \epsilon/[\zeta(2)+1]^{1/2}$, and choose $N$ such that $\sum_{i=N}^\infty 1/i^2 < \delta^2$, which exists since $|\zeta(2)| < \infty$. We claim that \begin{equation*} V = H \cap \left[\prod_{i=1}^{N-1} \left(x_i-\frac{\delta}{i},x_i+\frac{\delta}{i} ight) imes \prod_{i=N}^\infty \mathbb{R} ight] i \mathbf{x}, \end{equation*} basis element of the product topology is contained in $U$. Suppose $\mathbf{y} \in V$; then \begin{equation*} d(\mathbf{x},\mathbf{y})^2 = \sum_{i=1}^\infty (x_i-y_i)^2 < \delta^2 \sum_{i=1}^N \frac{1}{i^2} + \sum_{i=N}^\infty \frac{1}{i^2} < \delta^2[\zeta(2)+1] = \epsilon^2 \implies \mathbf{y} \in U, \end{equation*} i.e., $V \subseteq U \subseteq \mathcal{U}$, and product topology $\supset$ $\ell^2$-topology by Lemma $13.3$. Thus, we have the equality desired. \par It remains to show the box topology is strictly finer than the other topologies; since the other topologies are equal it suffices to show box topology $\supsetneq$ $\ell^2$-topology. But the open set $U = H \cap \prod (-1/i,1/i) i 0$ is open in the box topology but not the $\ell^2$-topology by the same argument as the proof that box topology $\supsetneq \ell^2$-topology in $(b)$, and so we are done. \end{proof}

Dan Whitman · Answer

\def\a{\alpha} \def\b{\beta} \def\d{\delta} \def\e{\epsilon} \def { ho} ewcommand{\vect}[1]{\mathbf{#1}} \def\va{\vect{a}} \def\vb{\vect{b}} ewcommand\parens[1]{\left( #1 ight)} ewcommand\seqfin[2]{\parens{#1_1, \ldots, #1_{#2}}} ewcommand\seqinf[1]{\parens{#1_1, #1_2, \ldots}} \def\br{\bar{ }} \def eals{{\mathbb{R}}} \def inf{ eals^\infty} \def\elltop{$\ell^2$-topology} \begin{lemma} \label{lem:metric:seqleq} Suppose that $\seqinf{a}$ and $\seqinf{b}$ are two real sequences that converge to $a$ and $b$, respectively. Then, if $a_n \leq b_n$ for every $n \in {{\mathbb{Z}}_+}$, then $a \leq b$. \end{lemma} \begin{proof} Suppose to the contrary that $a > b$, and let $\e = (a-b)/2$ so that clearly $\e > 0$. Since $\seqinf{a}$ converges to $a$ there is an $N_a \in {{\mathbb{Z}}_+}$ where $\left|a_n - a ight| < \e$ for all $n \geq N_a$. Similarly there is an $N_b \in {{\mathbb{Z}}_+}$ where $\left|b_n - b ight| < \e$ for all $n \geq N_b$ since $(b_n)$ converges to $b$. So let $N = \max\left\{N_a, N_b ight\}$ and consider any $n \geq N$. Then $n \geq N \geq N_a$ so that $\left|a_n - a ight| < \e$ and hence \begin{gather*} -\e < a_n - a < \e \ a - \e < a_n < a + \e \ a - \frac{a-b}{2} < a_n \ \frac{a+b}{2} < a_n \,. \end{gather*} Analogously, we have that $n \geq N \geq N_b$ so that $\left|b_n - b ight| < \e$ and so \begin{gather*} -\e < b_n - b < \e \ b - \e < b_n < b + \e \ b_n < b + \frac{a-b}{2} \ b_n < \frac{a+b}{2} \,. \end{gather*} Therefore we have $b_n < (a+b)/2 < a_n$, which contradicts the supposition that $b_n \geq a_n$. So it has to be that in fact $a \leq b$ as desired. \end{proof} \begin{corollary} \label{cor:metric:sumleq} Suppose that $\sum a_n$ and $\sum b_n$ are two real series that converge to $a$ and $b$, respectively. Then, if $a_n \leq b_n$ for every $n \in {{\mathbb{Z}}_+}$, then $a \leq b$. \end{corollary} \begin{proof} Since we have that $a_n \leq b_n$ for every $n \in {{\mathbb{Z}}_+}$, it follows that we have \begin{gather*} s_n = \sum_{i=1}^n a_i \leq \sum_{i=1}^n b_i = t_n \end{gather*} for any $n \in {{\mathbb{Z}}_+}$ for the partial sums. Then we have by definition of series that \begin{gather*} a = \sum_{n=1}^\infty a_n = \lim_{n o \infty} s_n \leq \lim_{n o \infty} t_n = \sum_{n=1}^\infty b_n = b \end{gather*} by Lemma~ ef{lem:metric:seqleq}, as desired. \end{proof} The following is a corollary of Lemma~\href{./exercise_20-4}{20.4.1}: \begin{corollary} \label{cor:metric:subball} Suppose that $X$ is a subspace of metric space $Y$ with metric $d$. If $U$ is open in the subspace topology on $X$ and contains $x$, then there is a ball $B_d(x, \e)$ in $Y$ such that $X \cap B_d(x,\e) \subset U$. \end{corollary} \begin{proof} Consider open $U$ in $X$ and any $x \in U$. Then there is an open set $V$ in $Y$ such that $U = X \cap V$ by the definition of the subspace topology. Then of course $x \in X$ and $x \in V$ since $x \in U$. It then follows that there is an $\e > 0$ such that $B_d(x, \e) \subset V$ by Lemma~\href{./exercise_20-4}{20.4.1}. Now consider any $y \in X \cap B_d(x, \e)$ so that $y \in X$ and $y \in B_d(x, \e)$. Then also $y \in V$ since $B_d(x, \e) \subset V$. Hence $y \in X \cap V = U$, which shows that $X \cap B_d(x, \e) \subset U$ as desired since $y$ was arbitrary. \end{proof} \begin{definition}\label{def:metric:pseries} If $\seqinf{x}$ is a sequence whose series $\sum x_i$ converges, we define the partial series starting at $n \in {{\mathbb{Z}}_+}$ as \begin{gather*} \sum_{i=n}^\infty x_i = \sum_{i=1}^\infty x_i - \sum_{i=1}^{n-1} x_i \,, \end{gather*} where we adopt the standard convention that $\sum_{i=a}^b x_i = 0$ when $b < a$. According to this the partial series starting at $n=1$ is just the series itself as expected. \end{definition} \begin{lemma} \label{lem:metric:pseries} If $\seqinf{x}$ is a series of non-negative real numbers such that the series $\sum x_i$ converges, then the sequence of partial series defined by \begin{gather*} p_n = \sum_{i=n}^\infty x_i \end{gather*} is non-increasing and converges to zero. Also each $p_n \geq 0$. \end{lemma} \begin{proof} Since the terms are all non-negative, clearly the sequence of partial sums is non-decreasing. Thus we have \begin{gather*} \sum_{i=1}^n x_i \leq \sum_{i=1}^{n+1} x_i \ -\sum_{i=1}^n x_i \geq -\sum_{i=1}^{n+1} x_i \ \sum_{i=1}^\infty x_i -\sum_{i=1}^n x_i \geq \sum_{i=1}^\infty x_i -\sum_{i=1}^{n+1} x_i \ \sum_{i=n+1}^\infty x_i \geq \sum_{i=n+2}^\infty x_i \ p_{n+1} \geq p_{n+2} \end{gather*} for any $n \in {{\mathbb{Z}}_+}$. Of course we also have \begin{gather*} 0 \leq x_1 = \sum_{i=1}^1 x_i \ 0 \geq -\sum_{i=1}^1 x_i \ \sum_{i=1}^\infty x_i \geq \sum_{i=1}^\infty x_i - \sum_{i=1}^1 x_i \ p_1 \geq p_2 \,. \end{gather*} Together these show that the sequence of partial series is non-increasing. Also, since the series of partial sums is non-decreasing, we have that that the infinite sum cannot be less than any of the partial sums, that is \begin{gather*} \sum_{i=1}^\infty x_i \geq \sum_{i=1}^{n-1} x_i \ \sum_{i=1}^\infty x_i - \sum_{i=1}^{n-1} x_i \geq 0 \ p_n \geq 0 \end{gather*} for any $n \in {{\mathbb{Z}}_+}$. To show that the sequence of partial series converges to zero, consider any $\e > 0$. We know that the sequence of partial sums converges to the infinite sum by the definition of a series. Hence there is an $N \in {{\mathbb{Z}}_+}$ such that \begin{align*} \left|\sum_{i=1}^n x_i - \sum_{i=1}^\infty x_i ight| &< \e \ \left|-p_{n+1} ight| &< \e \ \left|p_{n+1} ight| &< \e \ \left|p_{n+1} - 0 ight| &< \e \end{align*} for all $n \geq N$, and hence $\left|p_n - 0 ight| < \e$ for all $n \geq N+1$. This of course shows convergence to zero as desired. \end{proof} extbf{Main Problem.} (a) \begin{proof} First we show that the \elltop{} is finer than the topology inherited from the uniform topology using Lemma~20.2 since they are both metric topologies. So consider any $\mathbf{x} \in X$ and let $B = X \cap B_{\br}(\mathbf{x} , \e)$ for any $\e > 0$, which is of course a basis element of the subspace topology inherited by the uniform topology. Then the set $C = B_d(\mathbf{x} , \e/2)$ (where $d$ is the metric defined above for the \elltop{} instead of the usual metric on ${\mathbb{R}}$) is a basis element of the \elltop{}. We claim that $\mathbf{x} \in C \subset B$, which completes the proof that the \elltop{} is finer. First, it is obvious that $\mathbf{x} \in C$. Now consider any $\mathbf{y} \in C = B_d(\mathbf{x} , \e/2)$. Then we have that \begin{gather*} d(\mathbf{y} , \mathbf{x} ) = \left[\sum_{i=1}^\infty (y_i - x_i)^2 ight]^{1/2} < \e/2 \end{gather*} For $n \in {{\mathbb{Z}}_+}$ let \begin{gather*} s_n = \sum_{i=1}^n (y_i - x_i)^2 \end{gather*} be the partial sums of the infinite sum $\sum (y_i-x_i)^2$. Clearly each term in the sum is non-negative so that the sequence of partial sums is non-decreasing. It then follows that $s_n \leq \sum (y_i-x_i)^2$ for any $n \in {{\mathbb{Z}}_+}$. We clearly then have, for any $n \in {{\mathbb{Z}}_+}$, that \begin{gather*} \left|y_n-x_n ight|^2 = (y_n-x_n)^2 \leq \sum_{i=1}^{n-1} (y_i-x_i)^2 + (y_n-x_n)^2 = \sum_{i=1}^n (y_i-x_i)^2 = s_n \leq \sum_{i=1}^\infty (y_i-x_i)^2 \end{gather*} since again each term is non-negative. Hence by Corollary~\href{./exercise_20-1}{1} we have \begin{gather*} \left|y_n-x_n ight| = \left[\left|y_n-x_n ight|^2 ight]^{1/2} \leq \left[\sum_{i=1}^\infty (y_i-x_i)^2 ight]^{1/2} < \e/2 \,. \end{gather*} It then follows that \begin{gather*} \bar{p}(y_n,x_n) \leq p(y_n,x_n) = \left|y_n-x_n ight| < \e/2 \,, \end{gather*} where we have let $p$ and $\bar{p}$ denote the standard metric and standard bounded metric, respectively, on ${\mathbb{R}}$. Since this is true for any $n \in {{\mathbb{Z}}_+}$, it follows that \begin{gather*} \br(\mathbf{y} , \mathbf{x} ) = \sup\left\{\bar{p}(y_n,x_n) \mid n \in {{\mathbb{Z}}_+} ight\} \leq \e/2 < \e \end{gather*} so that $\mathbf{y} \in B_{\br}(\mathbf{x} , \e)$. Thus clearly $\mathbf{y} \in X \cap B_{\br}(\mathbf{x} , \e) = B$ so that $C \subset B$ as desired since $\mathbf{y} $ was arbitrary. Now we show that the topology inherited from the box topology is finer the \elltop{} using Lemma~13.3. So consider any $\mathbf{x} \in X$ and any basis element $B$ containing $\mathbf{x} $ of the \elltop{}. Then by Lemma~\href{./exercise_20-4}{20.4.1} there is an $\e > 0$ where $B_d(\mathbf{x} ,\e) \subset B$ since $B$ is of course open. Now consider the set \begin{gather*} C = X \cap \prod_{i \in {{\mathbb{Z}}_+}} B_p(x_i, \e/\sqrt{2^{i+1}}) \,, \end{gather*} where again $p$ denotes the usual metric on ${\mathbb{R}}$. Then clearly $C$ is a basis element of the subspace topology inherited by the box topology and contains $\mathbf{x} $, noting that clearly each $\e/\sqrt{2^{i+1}} > 0$. We claim that $C \subset B_d(\mathbf{x} , \e) \subset B$, which shows the desired result. To see this suppose that $\mathbf{y} \in C$ so that $\mathbf{y} \in X$ and $\mathbf{y} \in \prod B_p(x_i, \e/\sqrt{2^{i+1}})$. Then, for any $i \in {{\mathbb{Z}}_+}$, we have that \begin{gather*} p(y_i, x_i) = \left|y_i - x_i ight| \leq \e/\sqrt{2^{i+1}} \end{gather*} is true. It then follows from Lemma~\href{./exercise_20-1}{1} that \begin{gather*} (y_i - x_i)^2 = \left|y_i - x_i ight|^2 \leq \left(\e/\sqrt{2^{i+1}} ight)^2 \,. \end{gather*} Since this is true for any $i \in {{\mathbb{Z}}_+}$, we have by Corollary~ ef{cor:metric:sumleq} that \begin{align*} \sum_{i=1}^\infty (y_i-x_i)^2 &\leq \sum_{i=1}^\infty \left(\frac{\e}{\sqrt{2^{i+1}}} ight)^2 = \sum_{i=1}^\infty \frac{\e^2}{2^{i+1}} = \e^2 \sum_{i=1}^\infty \left(\frac{1}{2} ight)^{i+1} \ &= \e^2 \sum_{i=2}^\infty \left(\frac{1}{2} ight)^i = \e^2 \left[\sum_{i=0}^\infty \left(\frac{1}{2} ight)^i - \left(\frac{1}{2} ight)^0 - \left(\frac{1}{2} ight)^1 ight] \ &= \e^2 \left[\frac{1}{1-\frac{1}{2}} - 1 - \frac{1}{2} ight] = \e^2 \left[2 - 1 - \frac{1}{2} ight] \ &= \frac{\e^2}{2} \end{align*} since $\sum (1/2)^i$ is a geometric series. It then follows from Corollary~\href{./exercise_20-1}{1} that \begin{gather*} d(\mathbf{y} , \mathbf{x} ) = \left[\sum_{i=1}^\infty (y_i - x_i)^2 ight]^{1/2} \leq \left(\frac{\e^2}{2} ight)^{1/2} = \frac{\e}{\sqrt{2}} < \e \end{gather*} so that $\mathbf{y} \in B_d(\mathbf{x} , \e)$. Since $\mathbf{y} $ was arbitrary, this shows that $C \subset B_d(\mathbf{x} , \e) \subset B$ as desired, thereby completing the proof. \end{proof} (b) \begin{proof} Since relative topological ``fineness'' is preserved when inherited by subspace topologies (which is trivial to show), we have from part~(a) and what was shown in the text that \begin{gather*} ext{box topology} \supset ext{\elltop{}} \supset ext{uniform topology} \supset ext{product topology} \end{gather*} on $ inf$. To show that they are all distinct, it then suffices to show that each of the box, $\ell^2$, and uniform topologies (or more properly the subspace topologies inherited from them) have open sets that are not open in the $\ell^2$, uniform, and product topologies, respectively. First we show that the inherited box topology has an open set that is not open in the inherited \elltop{}. Consider the set $U = inf \cap \prod_{n \in {{\mathbb{Z}}_+}} (-1, 1/n)$, which is clearly a basis element and open set in the inherited box topology. We show that $U$ is not open in the inherited \elltop{} using the contrapositive of Corollary~ ef{cor:metric:subball}. So consider the point $\mathbf{0}$, which is clearly contained in $U$ and any $\e > 0$ and the arbitrary ball $B_d(\mathbf{0}, \e)$ of $X$. Of course, there is a positive integer $N$ large enough that $N \geq 2/\e$, and hence $1/N \leq \e/2$. Now, consider the sequence $\mathbf{x} $ defined by \begin{gather*} x_n = \begin{cases} 0 & n eq N \ \e/2 & n = N \end{cases} \end{gather*} for $n \in {{\mathbb{Z}}_+}$. Clearly $\mathbf{x} $ is eventually zero so that $\mathbf{x} \in inf$ since $x_n = 0$ for all $n \geq N+1$. We also clearly have \begin{gather*} d(\mathbf{x} , \mathbf{0}) = \left[\sum_{i=1}^\infty (x_i - 0)^2 ight]^{1/2} = \left[\sum_{i=1}^\infty x_i^2 ight]^{1/2} = \left[\left(\frac{\e}{2} ight)^2 ight]^{1/2} = \frac{\e}{2} < \e \end{gather*} so that $\mathbf{x} \in B_d(\mathbf{0}, \e)$. Therefore $\mathbf{x} \in inf \cap B_d(\mathbf{0}, \e)$. However, we have that $1/N \leq \e/2 = x_N$ so that $x_N otin (-1, 1/N)$, and hence $\mathbf{x} otin \prod_{n \in {{\mathbb{Z}}_+}} (-1, 1/n)$. From this clearly $\mathbf{x} otin U$ so that $ inf \cap B_d(\mathbf{0}, \e)$ cannot be a subset of $U$. Since the ball $B_d(\mathbf{0}, \e)$ was arbitrary, this shows that $U$ is not open in the inherited \elltop{} as desired. Next we show that there is an open set in the inherited \elltop{} that is not open in the inherited uniform topology. So consider the set $U = inf \cap B_d(\mathbf{0}, 1)$, which is clearly open in the inherited \elltop{}. We show that $U$ is not open in the inherited uniform topology, again by the contrapositive of Corollary~ ef{cor:metric:subball}. Consider the point $\mathbf{0}$, clearly in $U$, and any $\e > 0$ so that $B_{\br}(\mathbf{0}, \e)$ is an arbitrary ball in the uniform topology on ${\mathbb{R}}^{\omega}$. Now clearly there is an $N \in {{\mathbb{Z}}_+}$ large enough that $N \geq (2/\e)^2$. It then follows from Corollary~\href{./exercise_20-1}{1} that $\sqrt{N} \geq 2/\e$. Now define the sequence $\mathbf{x} $ by \begin{gather*} x_n = \begin{cases} \e/2 & n \leq N \ 0 & n > N \end{cases} \end{gather*} for $n \in {{\mathbb{Z}}_+}$, so that clearly $\mathbf{x} \in inf$. Then clearly we have \begin{gather*} \bar{p}(x_n, 0) \leq p(x_n, 0) = \left|\e/2 - 0 ight| = \left|\e/2 ight| = \e/2 \end{gather*} for any $n \leq N$, where again $p$ and $\bar{p}$ are the standard and standard bounded metrics on ${\mathbb{R}}$, respectively. If $n > N$ clearly \begin{gather*} \bar{p}(x_n, 0) \leq p(x_n, 0) = \left|0 - 0 ight| = \left|0 ight| = 0 \leq \e/2 \,. \end{gather*} Hence it follows that \begin{gather*} \br(\mathbf{x} , \mathbf{0}) = \sup\left\{\bar{p}(x_n, 0) \mid n \in {{\mathbb{Z}}_+} ight\} \leq \e/2 < \e \end{gather*} so that $\mathbf{x} \in B_{\br}(\mathbf{0}, \e)$. Therefore of course $\mathbf{x} \in inf \cap B_{\br}(\mathbf{0}, \e)$. However, we also have \begin{align*} d(\mathbf{x} , 0) &= \left[\sum_{i=1}^\infty (x_i - 0)^2 ight]^{1/2} = \left[\sum_{i=1}^\infty x_i^2 ight]^{1/2} = \left[\sum_{i=1}^N \left(\frac{\e}{2} ight)^2 ight]^{1/2} \ &= \left[N \left(\frac{\e}{2} ight)^2 ight]^{1/2} = \sqrt{N}\frac{\e}{2} \ &\geq \frac{2}{\e}\left(\frac{\e}{2} ight) = 1 \end{align*} since $\sqrt{N} \geq 2/\e$. Therefore clearly $\mathbf{x} otin B_d(\mathbf{0}, 1)$ so that $\mathbf{x} otin U$. It follows that $ inf \cap B_{\br}(\mathbf{0}, \e)$ cannot be a subset of $U$. Since $B_{\br}(\mathbf{0}, \e)$ was an arbitrary ball, this shows that $U$ is not open in the uniform topology as desired. Lastly, we show that there is an open set in the inherited uniform topology that is not open in the inherited product topology. So let $U = inf \cap B_{\br}(\mathbf{0}, 1)$, which is clearly open in the inherited uniform topology. We show that $U$ is not open in the inherited product topology using the definition of a basis. Consider any basis element $B$ of the inherited product topology that contains $\mathbf{0}$ so that $B = inf \cap \prod_{n \in {{\mathbb{Z}}_+}} B_n$, where each $B_n$ is open in ${\mathbb{R}}$ and $B_n eq {\mathbb{R}}$ for only finitely many $n \in {{\mathbb{Z}}_+}$. Then clearly there is an $N \in {{\mathbb{Z}}_+}$ where $B_N = {\mathbb{R}}$, and clearly we have $0 \in B_n$ for all $n \in {{\mathbb{Z}}_+}$. So define the sequence $\mathbf{x} $ by \begin{gather*} x_n = \begin{cases} 0 & n eq N \ 1 & n = N \end{cases} \end{gather*} for $n \in {{\mathbb{Z}}_+}$. Clearly $\mathbf{x} \in inf$ since $x_n = 0$ for all $n \geq N+1$. For any $n \in {{\mathbb{Z}}_+}$ we have that $x_n = 0 \in B_n$ if $n eq N$, and $x_n = 1 \in {\mathbb{R}} = B_n$ for $n = N$. Hence clearly $\mathbf{x} \in \prod B_n$ so that $\mathbf{x} \in inf \cap \prod B_n = B$ as well. However, we clearly have that \begin{gather*} p(x_N,0) = \left|x_N - 0 ight| = \left|1 - 0 ight| = \left|1 ight| = 1 \geq 1 \end{gather*} so that $\bar{p}(x_N, 0) = \min\left\{p(x_N,0), 1 ight\} = 1 \geq 1$. Then it has to be that \begin{gather*} \br(\mathbf{x} , \mathbf{0}) = \sup\left\{\bar{p}(x_n, 0) \mid n \in {{\mathbb{Z}}_+} ight\} \geq 1 \end{gather*} so that $\mathbf{x} otin B_{\br}(\mathbf{0}, 1)$ and hence $\mathbf{x} otin U$. This shows that $B$ is not a subset of $U$, which shows that $U$ is not open in the inherited product topology since $B$ was an arbitrary basis element. \end{proof} (c) First, we note that $H$ is contained in $X$ by the comparison test since the series $\sum (1/n)^2$ converges. Then, again since relative topological ``fineness'' is preserved when inherited by subspace topologies, we know that \begin{gather*} ext{box topology} \supset ext{\elltop{}} \supset ext{uniform topology} \supset ext{product topology} \end{gather*} on $H$. We claim, however, that the inherited box topology is distinct from the other three, which are all the same. \begin{proof} First, we show that the inherited box topology has an open set that is not open in the inherited \elltop{}, similar to how this was shown in part~(b). Consider the set $U = H \cap \prod_{n \in {{\mathbb{Z}}_+}} (-1, 1/n)$, which is clearly a basis element and open set in the inherited box topology. We show that $U$ is not open in the inherited \elltop{} using the contrapositive of Corollary~ ef{cor:metric:subball}. So consider the point $\mathbf{0}$, which is clearly contained in $U$ and any $\e > 0$ and the arbitrary ball $B_d(\mathbf{0}, \e)$ of $X$. Of course, there is positive integer $N$ large enough that $N \geq 2/\e$, and hence $1/N \leq \e/2$. Now, consider the sequence $\mathbf{x} $ defined by \begin{gather*} x_n = \begin{cases} 0 & n eq N \ 1/N & n = N \end{cases} \end{gather*} for $n \in {{\mathbb{Z}}_+}$. Clearly $\mathbf{x} \in H$ since each $x_n \in [0,1/n]$. We also clearly have \begin{gather*} d(\mathbf{x} , \mathbf{0}) = \left[\sum_{i=1}^\infty (x_i - 0)^2 ight]^{1/2} = \left[\sum_{i=1}^\infty x_i^2 ight]^{1/2} = \left[x_N^2 ight]^{1/2} = x_N = 1/N \leq \e/2 < \e \end{gather*} so that $\mathbf{x} \in B_d(\mathbf{0}, \e)$. Therefore $\mathbf{x} \in H \cap B_d(\mathbf{0}, \e)$. However, we have that $x_N = 1/N$ so that $x_N otin (-1, 1/N)$, and hence $\mathbf{x} otin \prod_{n \in {{\mathbb{Z}}_+}} (-1, 1/n)$. From this clearly $\mathbf{x} otin U$ so that $H \cap B_d(\mathbf{0}, \e)$ cannot be a subset of $U$. Since the ball $B_d(\mathbf{0}, \e)$ was arbitrary, this shows that $U$ is not open in the inherited \elltop{} as desired. To show that the other three topologies are the same on $H$, it suffices to show that the inherited product topology is finer than the inherited \elltop{}, which we do using Lemma~13.3. So consider any $\mathbf{x} \in H$ and any basis element $B$ of the inherited \elltop{} that contains $\mathbf{x} $. It then follows from Corollary~ ef{cor:metric:subball} that there is an $\e > 0$ where $B' = H \cap B_d(\mathbf{x} , \e) \subset B$ since of course $B$ is open. Now, by Lemma~ ef{lem:metric:pseries} the sequence of partial series $p_n = \sum_{i=n}^\infty (1/i)^2$ converges to zero so that there is an $N \in {{\mathbb{Z}}_+}$ where $p_n = \left|p_n ight| = \left|p_n - 0 ight| < \e^2/2$ for all $n \geq N$ since each $p_n$ is non-negative (also by Lemma~ ef{lem:metric:pseries}). In particular $p_{N+1} = \sum_{i=N+1}^\infty (1/i)^2 < \e^2/2$. So define the following sets: \begin{gather*} C_n = \begin{cases} B_p(x_n, \e/\sqrt{2N}) & n \leq N \ {\mathbb{R}} & n > 0 \end{cases} \end{gather*} for $n \in {{\mathbb{Z}}_+}$, where again $p$ is the usual metric on ${\mathbb{R}}$. Clearly $C = H \cap \prod C_n$ is a basis element in the inherited product topology that contains $\mathbf{x} $. We now claim that $C \subset B' \subset B$, which shows that the inherited product topology is finer by Lemma~13.3 since $B$ was an arbitrary basis element. To see this, consider any $\mathbf{y} \in C$ so that $\mathbf{y} \in H$ and $\mathbf{y} \in \prod C_n$. Now, for any $n \leq N$, we have that $y_n \in C_n = B_p(x_n, \e/\sqrt{2N})$ so that $\left|y_n - x_n ight| < \e/\sqrt{2N}$. It then follows from Lemma~\href{./exercise_20-1}{1} that \begin{gather*} (y_n - x_n)^2 = \left|y_n - x_n ight|^2 < \left(\frac{\e}{\sqrt{2N}} ight)^2 = \frac{\e^2}{2N} \,. \end{gather*} Since this is true of each $n \leq N$, we clearly have that \begin{gather*} \sum_{i=1}^N (y_i - x_i)^2 < \sum_{i=1}^N \frac{\e^2}{2N} = N\frac{\e^2}{2N} = \frac{\e^2}{2} \,. \end{gather*} Now, for any $n > N$ we have that of course that $\mathbf{y} \in H = \prod [0, 1/n]$ so that $y_n \in [0, 1/n]$. Similarly $\mathbf{x} \in H$ so that $x_n \in [0,1/n]$. Then both $0 \leq y_n \leq 1/n$ and $0 \leq x_n \leq 1/n$ so that \begin{gather*} 0 \leq x_n \leq 1/n \ 0 \geq -x_n \geq -1/n \ y_n \geq y_n - x_n \geq y_n - 1/n \ 1/n \geq y_n \geq y_n - x_n \geq y_n - 1/n \geq 0 - 1/n = -1/n \,. \end{gather*} Hence $\left|y_n - x_n ight| \leq 1/n$, from which it follows that $(y_n-x_n)^2 = \left|y_n-x_n ight|^2 \leq (1/n)^2$ by Lemma~\href{./exercise_20-1}{1}. Since this it true for any $n > N$, it follow from either Lemma~ ef{lem:metric:seqleq} or Corollary~ ef{cor:metric:sumleq} that \begin{gather*} \sum_{i=N+1}^\infty (y_i - x_i)^2 \leq \sum_{i=N+1}^\infty \left(\frac{1}{i} ight)^2 < \frac{\e^2}{2} \,. \end{gather*} We then have, from the definition of partial series (Definition~ ef{def:metric:pseries}), that \begin{gather*} \sum_{i=1}^\infty (y_i - x_i)^2 = \sum_{i=1}^N (y_i - x_i)^2 + \sum_{i=N+1}^\infty (y_i - x_i)^2 < \frac{\e^2}{2} + \frac{\e^2}{2} = \e^2 \end{gather*} Then Corollary~\href{./exercise_20-1}{1} means that \begin{gather*} d(\mathbf{y} , \mathbf{x} ) = \left[\sum_{i=1}^\infty (y_i - x_i)^2 ight]^{1/2} < \left(\e^2 ight)^{1/2} = \e \end{gather*} so that $\mathbf{y} \in B_d(\mathbf{x} , \e)$, and hence clearly $\mathbf{y} \in H \cap B_d(\mathbf{x} , \e) = B'$. Since $\mathbf{y} $ was arbitrary, this shows that $C \subset B' \subset B$ as desired. \end{proof}

Exercise 20.8

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Exercise 20.8

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