Exercise 20.6

Question

Let $\overline{ho}$ be the uniform metric on $\mathbb{R}^\omega$. Given $\mathbf{x} = (x_1,x_2,\ldots) \in \mathbb{R}^\omega$ and given $0 < \epsilon < 1$, let
\begin{equation*}
  U(\mathbf{x},\epsilon) = (x_1-\epsilon,x_1+\epsilon) 	imes \cdots 	imes (x_n-\epsilon,x_n+\epsilon) 	imes \cdots .
\end{equation*}
\begin{enumerate}[label=(\alph*)]
\item Show that $U(\mathbf{x},\epsilon)$ is not equal to the $\epsilon$-ball $B_{\overline{ho}}(\mathbf{x},\epsilon)$.
\item Show that $U(\mathbf{x},\epsilon)$ is not even open in the uniform topology.
\item Show that
  \begin{equation*}
    B_{\overline{ho}}(\mathbf{x},\epsilon) = \bigcup_{\delta < \epsilon} U(\mathbf{x},\delta).
  \end{equation*}
\end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $(a)$] Consider the point \begin{equation*} \mathbf{y} = (y_n)_{n \in \mathbb{N}}, \quad y_n = x_n + \epsilon \sum_{k=1}^n \frac{1}{2^k} = x_n + \epsilon\left( 1 - \frac{1}{2^n} ight). \end{equation*} $y_n \in (x_n-\epsilon,x_n+\epsilon)$ for all $n$ implies $\mathbf{y} \in U(\mathbf{x},\epsilon)$, while $\mathbf{y} otin B_{\overline{ ho}}(\mathbf{x},\epsilon)$ since \begin{equation*} \overline{ ho}(\mathbf{x},\mathbf{y}) = \sup_n \overline{d}(x_n,y_n) = \epsilon. \end{equation*} \end{proof} \begin{proof}[Proof of $(b)$] $U(\mathbf{x},\epsilon)$ is not open since the point $\mathbf{y}$ in $(a)$ has no neighborhood contained in $B_{\overline{ ho}}(\mathbf{x},\epsilon)$. For, suppose $B_{\overline{ ho}}(\mathbf{y},\delta) \subseteq U(\mathbf{x},\epsilon)$. We can find $N$ such that \begin{equation*} \frac{\delta}{2} > \epsilon \sum_{k=N+1}^\infty \frac{1}{2^k}, \end{equation*} since $\sum 1/2^k$ converges, and so its tail becomes infinitesimally small. We see that then, defining $\mathbf{y}'$ such that $y'_n = y_n$ for all $n e N$ and $y'_N = y_N + \delta/2$, $\mathbf{y}' \in B_{\overline{ ho}}(\mathbf{y},\delta)$ but $\mathbf{y}' otin U(\mathbf{x},\epsilon)$ since $y'_n = y_n + \delta/2 > y_n + \epsilon \sum_{k=N+1}^\infty \frac{1}{2^k} = x_n + \epsilon$, a contradiction. \end{proof} \begin{proof}[Proof of $(c)$] The $\supseteq$ direction is clear, since each $U(\mathbf{x},\delta) \subseteq B_{\overline{ ho}}(\mathbf{x},\epsilon)$ by the fact that $\delta < \epsilon$. Now suppose $\mathbf{z} \in B_{\overline{ ho}}(\mathbf{x},\epsilon)$; if $\overline{ ho}(\mathbf{x},\mathbf{z}) = \xi$, then we can find $\delta \in (\xi,\epsilon)$ so that $\mathbf{z} \in U(\mathbf{x},\xi)$, i.e., $B_{\overline{ ho}}(\mathbf{x},\epsilon) \subseteq \bigcup_{\delta < \epsilon} U(\mathbf{x},\delta)$. \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} (a) \begin{proof} We show that $B_{\br}(\mathbf{x} , \e)$ is not a subset of $U(\mathbf{x} , \e)$, which of course suffices to show that they cannot be equal. To this end we define the point $\mathbf{y} $ in ${\mathbb{R}}^{\omega}$ by \begin{gather*} y_n = x_n + \e\left(1 - \frac{1}{n} ight) \end{gather*} for any $n \in {{\mathbb{Z}}_+}$. Then, for any such $n$, we clearly have that \begin{align*} n &\geq 1 \ -n &\leq -1 < 0 \ -1 &\leq -\frac{1}{n} < 0 & ext{(since $n > 0$)} \ 0 &\leq 1 - \frac{1}{n} < 1 \ 0 &\leq \e\left(1 - \frac{1}{n} ight) < \e & ext{(since $\e > 0$)} \ x_n &\leq x_n + \e\left(1 - \frac{1}{n} ight) < x_n + \e \ x_n - \e < x_n &\leq y_n < x_n + \e \end{align*} so that $y_n \in (x_n - \e, x_n + \e)$. Since $n$ was arbitrary, this shows that $\mathbf{y} \in \prod_{n=1}^\infty (x_n - \e, x_n + \e) = U(\mathbf{x} , \e)$. However, again for any $n \in {{\mathbb{Z}}_+}$, it was shown that $x_n \leq y_n$ so that \begin{gather*} d(y_n, x_n) = \left|y_n - x_n ight| = y_n - x_n = x_n + \e\left(1 - \frac{1}{n} ight) - x_n = \e\left(1 - \frac{1}{n} ight) \,. \end{gather*} It was shown above that \begin{align*} 0 \leq \e\left(1 - \frac{1}{n} ight) < \e &< 1 \ 0 \leq d(y_n, x_n) < \e &< 1 \end{align*} so that \begin{gather*} \bar{d}(y_n, x_n) = \min\left\{d(y_n, x_n), 1 ight\} = d(y_n, x_n) \,. \end{gather*} We then clearly have \begin{gather*} \lim_{n o \infty} \bar{d}(y_n, x_n) = \lim_{n o \infty} d(y_n, x_n) = \lim_{n o \infty} \e \left(1 - \frac{1}{n} ight) = \e \end{gather*} so that \begin{gather*} \br(\mathbf{y} , \mathbf{x} ) = \sup_{n \in {{\mathbb{Z}}_+}} \bar{d}(y_n, x_n) = \e \geq \e \end{gather*} since the sequence $\mathbf{y} $ is clearly monotonically increasing. This shows that $\mathbf{y} otin B_{\br}(\mathbf{x} , \e)$ so that $B_{\br}(\mathbf{x} , \e)$ cannot be a subset of $U(\mathbf{x} , \e)$. \end{proof} (b) \begin{proof} Let $\mathbf{y} $ be the point in ${\mathbb{R}}^{\omega}$ defined in part~(a) so that we know that $\mathbf{y} \in U(\mathbf{x} , \e)$. Now if $U(\mathbf{x} , \e)$ were open in the uniform topology then there would be a basis element $B_{\br}(\mathbf{y} , \d)$ that is contained in $U(\mathbf{x} , \e)$. We shall show that any such basis element cannot be contained within $U(\mathbf{x} , \e)$, from which the desired result follows. So consider any $\d > 0$ so that there is an $n \in {{\mathbb{Z}}_+}$ large enough that $n > \e/\d$. Then we have \begin{align*} n &> \frac{\e}{\d} \ \d &> \e \frac{1}{n} & ext{(since both $\d > 0$ and $n > 0$)} \ -\e\frac{1}{n} + \d &> 0 \ \e - \e\frac{1}{n} + \d &> \e \ x_n + \e\left(1 - \frac{1}{n} ight) + \d &> x_n + \e \ y_n + \d &> x_n + \e \,. \end{align*} Now define the point $\mathbf{z} $ by \begin{gather*} z_m = \begin{cases} y_m & m eq n \ \frac{(x_n + \e) + (y_n + \d)}{2} & m = n \,. \end{cases} \end{gather*} It then follows that $x_n + \e < z_n < y_n + \d$. The fact that $x_n + \e < z_n$ means that of course $z_n otin (x_n - \e, x_n + \e)$ so that $\mathbf{z} otin \prod_{m=1}^\infty (x_m - \e, x_m + \e) = U(\mathbf{x} , \e)$. However, for $m eq n$ we have that \begin{gather*} d(z_m, y_m) = d(y_m, y_m) = 0 \end{gather*} and so $\bar{d}(z_m, y_m) = 0$ as well. For $m = n$ we have \begin{gather*} x_m + \e = x_n + \e < z_m = z_n < y_n + \d = y_m + \d \ x_m + \e - y_m < z_m - y_m < \d \ x_m + \e - x_m - \e\left(1 - \frac{1}{n} ight) < z_m - y_m < \d \ 0 < \e\frac{1}{n} < z_m - y_m < \d \ \left|z_m - y_m ight| < \d \ \bar{d}(z_m, y_m) \leq d(z_m, y_m) < \d \,. \end{gather*} From these facts it follows that \begin{gather*} \br(\mathbf{z} , \mathbf{y} ) = \sup_{m \in {{\mathbb{Z}}_+}} \bar{d}(z_m, y_m) = \bar{d}(z_n, y_n) < \d \end{gather*} so that $\mathbf{z} \in B_{\br}(\mathbf{y} , \d)$. This shows that $B_{\br}(\mathbf{y} , \d)$ is not a subset of $U(\mathbf{x} , \e)$, which shows the desired result as explained before. \end{proof} (c) \begin{proof} $(\subset)$ Let $\mathbf{y} $ be any element of $B_{\br}(\mathbf{x} , \e)$ so that $\br(\mathbf{y} , \mathbf{x} ) < \e$. Then there is a $\d$ where $\br(\mathbf{y} , \mathbf{x} ) < \d < \e$ since the reals are order-dense. For any $n \in {{\mathbb{Z}}_+}$ it must be that $\bar{d}(y_n, x_n) < \d < \e < 1$ since $\br$ is the supremum of these. From this it has to be that $\bar{d}(y_n, x_n) = d(y_n, x_n) < \d$ so that \begin{gather*} d(y_n, x_n) = \left|y_n - x_n ight| < \d \ -\d < y_n - x_n < \d \ x_n - \d < y_n < x_n + \d \,. \end{gather*} Hence $y_n \in (x_n - \d, x_n + \d)$. Since $n$ was arbitrary, this shows that $\mathbf{y} \in \prod_{n=1}^\infty (x_n-\d, x_n+\d) = U(\mathbf{x} , \d)$. Thus obviously $\mathbf{y} \in \bigcup_{\d < \e} U(\mathbf{x} , \d)$, which shows the desired result. $(\supset)$ Now suppose that $\mathbf{y} \in \bigcup_{\d < \e} U(\mathbf{x} , \d)$ so that there is a $\d < \e$ where $\mathbf{y} \in U(\mathbf{x} , \d)$. Consider any $n \in {{\mathbb{Z}}_+}$ so that we have $y_n \in (x_n-\d, x_n+\d)$. Then of course \begin{gather*} x_n - \d < y_n < x_n + \d \ -\d < y_n - x_n < \d \end{gather*} so that $d(y_n, x_n) = \left|y_n - x_n ight| < \d < \e < 1$ so that it must be that $\bar{d}(y_n, x_n) = d(y_n, x_n) < \d$. Since $n$ was arbitrary, it follows that \begin{gather*} \br(\mathbf{y} , \mathbf{x} ) = \sup\left\{\bar{d}(y_n, x_n) \mid n \in {{\mathbb{Z}}_+} ight\} \leq \d < \e \,, \end{gather*} and hence $\mathbf{y} \in B_{\br}(\mathbf{x} , \e)$. Since $\mathbf{y} $ was arbitrary, this shows that $B_{\br}(\mathbf{x} , \e) \supset \bigcup_{\d < \e} U(\mathbf{x} , \d)$, which completes the proof. \end{proof}

Exercise 20.6

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Exercise 20.6

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