Exercise 20.7

Question

Consider the map $h: {\mathbb{R}}^{\omega} 	o {\mathbb{R}}^{\omega}$ defined in \href{./exercise_19-8}{Exercise~8} of $\S 19$; give ${\mathbb{R}}^{\omega}$ the uniform topology.
Under what conditions on the numbers $a_i$ and $b_i$ is $h$ continuous?
a homeomorphism?

Dan Whitman · Accepted 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} First some discussion. We know that the product topology is strictly finer than the uniform topology in ${\mathbb{R}}^{\omega}$, and that box topology is strictly finer than the uniform topology. By Lemma~\href{./exercise_20-4}{20.4.2} part~(1), when the topology on the range of a function becomes coarser, the function remains continuous. It is similarly easy to show that if the topology on the \emph{domain} of a function becomes \emph{finer}, it also remains continuous. However, nothing can be said for sure if the range becomes finer and/or the domain becomes coarser. It was shown in \href{./exercise_19-8}{Exercise~19.8} that $h$ is a homeomorphism (for $a_i > 0$ as in the exercise) if both the domain and range have the product topology, or if they both have the box topology. By what was just discussed then, $h$ is at least continuous with box topology on the domain, and the uniform topology on the range, or likewise with the uniform topology on the domain and the product topology on the range. However, the relative ``fineness'' of these topologies does not allow us to conclude anything about whether $h$ is continuous or a homeomorphism when both the domain and range are the uniform topologies, which is unfortunately what we are interested in. In fact, we claim that $h$ is continuous with the uniform topology as the domain and range if and only if the set of numbers $\left\{a_i ight\}_{i \in {{\mathbb{Z}}_+}}$ is bounded (and of course each $a_i > 0$). \begin{proof} $(\Rightarrow)$ We show this direction by contrapositive. So suppose that $\left\{a_i ight\}$ is not bounded. We then show that $h$ is not continuous by showing the negation of Theorem~18.1 part~(4). So consider the point $\mathbf{0}$ and the neighborhood $V = B_{\br}(h(\mathbf{0}), 1)$ in the uniform topology. Consider also any neighborhood $U$ of $\mathbf{0}$ so that by Lemma~\href{./exercise_20-4}{20.4.1} there is an $\e > 0$ where $B_{\br}(\mathbf{0}, \e) \subset U$. Now define the point $\mathbf{x} \in {\mathbb{R}}^{\omega}$ by $x_i = \e/2$ for all $i \in {{\mathbb{Z}}_+}$. Then, for any $i \in {{\mathbb{Z}}_+}$, we have \begin{gather*} \bar{d}(x_i, 0) \leq d(x_i, 0) = \left|x_i - 0 ight| = \left|x_i ight| = \left|\e/2 ight| = \e/2 \end{gather*} so that clearly \begin{gather*} \br(\mathbf{x} , \mathbf{0}) = \sup\left\{\bar{d}(x_i, 0) \mid i \in {{\mathbb{Z}}_+} ight\} \leq \e/2 < \e \,, \end{gather*} and hence $\mathbf{x} \in B_{\br}(\mathbf{0}, \e)$ so that also $\mathbf{x} \in U$. Then clearly $h(\mathbf{x} ) \in h(U)$. Now, since the $a_i$ coefficients are unbounded, there is a specific $i \in {{\mathbb{Z}}_+}$, where $a_i \geq 2/\e$ regardless of how small $\e$ is. We then have that \begin{gather*} d(h_i(x_i), h_i(0)) = d(a_i x_i + b_i, b_i) = \left|a_i x_i + b_i - b_i ight| = \left|a_i x_i ight| = a_i \left|x_i ight| = a_i \frac{\e}{2} \geq \frac{2}{\e} \frac{\e}{2} = 1 \,, \end{gather*} from which we have $\bar{d}(h_i(x_i), h_i(0)) = 1$ and so \begin{gather*} \br(h(\mathbf{x} ), h(\mathbf{0})) = \sup\left\{\bar{d}(h_i(x_i), h_i(0)) \mid i \in {{\mathbb{Z}}_+} ight\} \geq 1 \,. \end{gather*} Then of course $h(\mathbf{x} ) otin B_{\br}(h(\mathbf{0}), 1) = V$. This shows that $h(U) ot \subset V$, which in turn shows that $h$ is not continuous, since $U$ was an arbitrary neighborhood of $\mathbf{0}$. $(\Leftarrow)$ Now suppose that the coefficients $a_i$ are bounded so that there is a real $a$ where $0 < a_i \leq a$ for all $i \in {{\mathbb{Z}}_+}$. Consider any $\mathbf{x} \in {\mathbb{R}}^{\omega}$ and any neighborhood $V$ of $h(\mathbf{x} )$ in the uniform topology. Then there is an $\e > 0$ where $B_{\br}(h(\mathbf{x} ), \e) \subset V$ by Lemma~\href{./exercise_20-4}{20.4.1}. So let $\d = \min\left\{\e/2a, 1 ight\}$, noting that $\d > 0$ since both $\e > 0$ and $a > 0$. Then $U = B_{\br}(\mathbf{x} , \d)$ is of course a neighborhood of $\mathbf{x} $ in the uniform topology. We claim that $h(U) \subset V$. To see this, suppose that $\mathbf{z} \in h(U)$ so that there is a $\mathbf{y} \in U$ where $\mathbf{z} = h(\mathbf{y} )$. Then $\br(\mathbf{y} , \mathbf{x} ) < \d \leq 1$ since $\mathbf{y} \in U$, from which it follows that each $\bar{d}(y_i, x_i) < \d \leq 1$, and hence \begin{gather*} \bar{d}(y_i, x_i) = d(y_i, x_i) = \left|y_i - x_i ight| < \d \,. \end{gather*} We then have that \begin{align*} \bar{d}(h_i(y_i), h_i(x_i)) &\leq d(h_i(y_i), h_i(x_i)) = d(a_i y_i + b_i, a_i x_i + b_i) = \left|a_i y_i + b_i - a_i x_i - b_i ight| \ &= \left|a_i y_i - a_i x_i ight| = \left|a_i(y_i - x_i) ight| = a_i \left|y_i - x_i ight| \leq a \left|y_i - x_i ight| \ &< a\d \leq a \e/2a = \e/2 \end{align*} for each $i \in {{\mathbb{Z}}_+}$. From this, it follows that \begin{gather*} \br(\mathbf{z} , h(\mathbf{x} )) = \br(h(\mathbf{y} ), h(\mathbf{x} )) = \sup\left\{\bar{d}(h_i(y_i), h_i(x_i)) \mid i \in {{\mathbb{Z}}_+} ight\} \leq \e/2 < \e \end{gather*} so that $\mathbf{z} \in B_{\br}(h(\mathbf{x} ), \e) \subset V$. Since $\mathbf{z} $ was arbitrary, this shows that $f(U) \subset V$, which shows that $h$ is continuous by Theorem~18.1 part~(4) since $V$ was an arbitrary neighborhood. \end{proof} The function $h$ is a homeomorphism if and only if there are real $a, a_0 > 0$ where $0 < a_0 \leq a_i \leq a$ for all $i \in {{\mathbb{Z}}_+}$. \begin{proof} First, it was just shown that $h$ is continuous if and only if $\left\{a_i ight\}$ is bounded above. It was shown in Exercise~19.8 that $h$ is bijective (so long as each $a_i > 0$) and that its inverse function ${h}^{-1}$ has the same form as $h$: \begin{gather*} {h}^{-1}(\mathbf{y} ) = (c_i y_i + d_i)_{i \in {{\mathbb{Z}}_+}} \,, \end{gather*} where each $c_i = 1/a_i$ and $d_i = -b_i/a_i$. Since the topologies of the domain and range of ${h}^{-1}$ are both the uniform topology, as with $h$, it follows that the same conditions on $c_i$ and $d_i$ will make ${h}^{-1}$ continuous. That is to say that ${h}^{-1}$ is continuous (and thus $h$ is a homeomorphism) if and only if also $\left\{c_i ight\} = \left\{1/a_i ight\}$ is bounded above. Of course, $\left\{1/a_i ight\}$ being bounded above means that $\left\{a_i ight\}$ cannot get arbitrarily close to zero and so must have some nonzero lower bound $a_0$. \end{proof}

Exercise 20.7

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Exercise 20.7

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