# Distinctions of different topologies on the sequence space (countable cartesian products of $\mathbb{R}$)


Question

I solved this exercise in Munkres.(20.4) But I don't know if I did it righ t or not. I really appreciate it if anyone can take a look at it and give me some pointers.

Problem

Consider the product, uniform, and box topologies on $\R^{\omega}$, where $\R^\omega$ is defined to the cartesian product of countably infinite copies of $\R$.

(1). In which topologies are the following functions from $\R$ to $\R^\omega$ continuous?

\begin{aligned} f(t)&=(t,2t,3t,...),\\ g(t)&=(t,t,t,...),\\ h(t)&=(t,\frac{1}{2}t,\frac{1}{3}t,...). \end{aligned}

(2). In which topologies do the following sequences converge?

\begin{aligned} \b{w}_1&=(1,1,1,1,...),\b{x}_1=(1,1,1,1,...),\\ \b{w}_2&=(0,2,2,2,...),\b{x}_2=(0,\frac{1}{2},\frac{1}{2},\frac{1}{2},...),\\ \b{w}_3&=(0,0,3,3,...),\b{x}_3=(0,0,\frac{1}{3},\frac{1}{3},...),\\ &\cdots, \cdots\\ \b{y}_1&=(1,0,0,0,...),\b{z}_1=(1,1,0,0,...),\\ \b{y}_2&=(\frac{1}{2},\frac{1}{2},0,0,...),\b{z}_2=(\frac{1}{2},\frac{1}{2},0,0,...),\\ \b{y}_3&=(\frac{1}{3},\frac{1}{3},\frac{1}{3},0,...),\b{z}_3=(\frac{1}{3},\frac{1}{3},0,0,...),\\ &\cdots,\cdots \end{aligned}

Solution

(a).

We divide our solution in separate cases.

In the $\textbf{product}$ $\textbf{topology}$, $f\left(t\right)$, $g\left(t\right)$ and $h\left(t\right)$ are all continuous.

For $f\left(t\right)$, since $\pi_{n}\circ f\left(t\right)=nt$ is continuous for all $n\in\N,$ in view of Thm. 19.6 we conclude that $f$ is continuous. Similarly, we claim $g\left(t\right)$ and $h\left(t\right)$ to be continuous for the same reason.

In the $\textbf{box topology}$, none of $f\left(t\right)$ and $g\left(t\right)$ and $h\left(t\right)$ are continuous.

Let $B=\prod_{n=1}^{\infty}B_{n}$ where $B_{n}=\left(-1/n^{2},1/n^{2}\right)$. Suppose $f$ is continuous; then $f^{-1}\left(B\right)$ is open. Since $f\left(0\right)=0\in B$, we have $0\in f^{-1}\left(B\right)$ , which by assumption is open; hence there exists some $\delta>0$ such that $0\in\left(-\delta,\delta\right)\subset f^{-1}\left(B\right)$. But \begin{align} \left(-\delta,\delta\right)\subset f^{-1}\left(B\right)\implies & f\left(\left(-\delta,\delta\right)\right)\subset\prod_{n=1}^{\infty}\left(-\frac{1}{n^{2}},\frac{1}{n^{2}}\right),\\ \implies & \left(-n\delta,n\delta\right)\subset\left(-\frac{1}{n^{2}},\frac{1}{n^{2}}\right)\mbox{ for all }n,\\ \implies & \left(-\delta,\delta\right)\subset\left(-\frac{1}{n^{3}},\frac{1}{n^{3}}\right)\mbox{ for all }n, \end{align} which is a contradiction since we can always find sufficiently large $n$ such that $\left(-\delta,\delta\right)\nsubseteq\left(-1/n^{3},1/n^{3}\right)$. Therefore, $f$ is not continuous. Using the same line, we can show that $g\left(t\right)$ and $h\left(t\right)$ are not continuous as well.

In the $\textbf{uniform topology}$, we assert that $g\left(t\right)$ and $h\left(t\right)$ are continuous but $f\left(t\right)$ is not continuous.

We first show that $g$ and $h$ are continuous. In view of Theorem 18.2-e, it suffices for us to show that $g\left(t\right)$ is continuous from $\R$ to its image, $g\left(\R\right).$ Pick an aribitrary point $g\left(x_{0}\right)\in g\left(\R\right)$. We have \begin{align} B_{\bar{\rho}}\left(g\left(x_{0}\right),\varepsilon\right)= & \left\{ x\in\R\vert\sup_{i\in\N}\left\{ \bar{d}_{i}\left(g\left(x\right),g\left(x_{0}\right)\right)\right\} <\varepsilon\right\} . \end{align} Since $x_{0}\in g^{-1}\left(B_{\bar{\rho}}\left(g\left(x_{0}\right),\varepsilon\right)\right)$, it suffices for us a find an open set $U$ such that $x_{0}\in U$ and $U\in g^{-1}\left(B_{\bar{\rho}}\left(g\left(x_{0}\right),\varepsilon\right)\right)$. We define such $U$ to be \begin{align} U= & \left(x_{0}-\frac{1}{2}\varepsilon,x_{0}+\frac{1}{2}\varepsilon\right). \end{align} That $x_{0}\in U$ is trivial by construction. Since $g\left(U\right)=\prod_{i=1}^{\infty}U$ and for any $\mathbf{y}\in\prod_{i=1}^{\infty}U$ \begin{align} \bar{\rho}\left(\mathbf{y},g\left(x_{0}\right)\right)= & \sup_{i\in\N}\left\{ \bar{d}_{i}\left(y_{i},x_{0}\right)\right\} \leq\frac{1}{2}\varepsilon<\varepsilon, \end{align} this implies $\mathbf{y}\in B_{\bar{\rho}}\left(g\left(x_{0}\right),\varepsilon\right)$; as $\mathbf{y}$ is arbitrary, we have $g\left(U\right)\subset B_{\bar{\rho}}\left(g\left(x_{0}\right),\varepsilon\right)$ which in turn implies $U\subset g^{-1}\left(B_{\bar{\rho}}\left(g\left(x_{0}\right),\varepsilon\right)\right)$ as desired; hence $g\left(t\right)$ is continuous.

We use the same technique to show that $h$ is continuous. Pick an aribitrary point $g\left(x_{0}\right)\in g\left(\R\right)$. We have \begin{align} B_{\bar{\rho}}\left(h\left(x_{0}\right),\varepsilon\right)= & \left\{ x\in\R\vert\sup_{i\in\N}\left\{ \bar{d}_{i}\left(h\left(x\right),h\left(x_{0}\right)\right)\right\} <\varepsilon\right\} . \end{align} Since $x_{0}\in h^{-1}\left(B_{\bar{\rho}}\left(h\left(x_{0}\right),\varepsilon\right)\right)$, it suffices for us a find an open set $U$ such that $x_{0}\in U$ and $U\in h^{-1}\left(B_{\bar{\rho}}\left(h\left(x_{0}\right),\varepsilon\right)\right)$. We define such $U$ to be \begin{align} U= & \left(x_{0}-\frac{1}{2}\varepsilon,x_{0}+\frac{1}{2}\varepsilon\right). \end{align} That $x_{0}$ is in $U$ is obvious by construction. Notice that $g\left(U\right)=\prod_{n=1}^{\infty}B_{n}$where \begin{align} B_{n}= & \left(-\frac{1}{n}\left(x_{0}-\frac{1}{2}\varepsilon\right),\frac{1}{n}\left(x_{0}+\frac{1}{2}\varepsilon\right)\right). \end{align} Let $\mathbf{y}$ in arbitrary in $h\left(U\right)$; then we have \begin{align} \bar{\rho}\left(\mathbf{y},h\left(x_{0}\right)\right)= & \sup_{i\in\N}\left\{ \bar{d_{i}}\left(y_{i},h_{i}\left(x_{0}\right)\right)\right\} \\ = & \sup_{i\in\N}\left\{ \bar{d}_{i}\left(y_{i},\frac{1}{i}x_{0}\right)\right\} \leq\frac{\varepsilon}{2}<\varepsilon. \end{align} Hence $\mathbf{y}\in B_{\bar{\rho}}\left(g\left(x_{0}\right),\varepsilon\right)$. As $\mathbf{y}$ is arbitrary, $g\left(U\right)\subset B_{\bar{\rho}}\left(g\left(x_{0}\right),\varepsilon\right)$ and as a result, $U\subset h^{-1}\left(B_{\bar{\rho}}\left(g\left(x_{0}\right),\varepsilon\right)\right)$ as desired; thus $h\left(t\right)$ is continuous.

To show that $f$ is not continuous. Pick any $f\left(x_{0}\right)$ in $f\left(\R\right)$. For arbitrary $0<\varepsilon<1$, we have \begin{align} f^{-1}\left(B_{\bar{\rho}}\left(f\left(x_{0}\right),\varepsilon\right)\right)= & \left\{ x\in\R\vert\bar{\rho}\left(f\left(x\right),f\left(x_{0}\right)\right)<\varepsilon\right\} \\ = & \left\{ x\in\R\vert\sup_{i\in\N}\left\{ \bar{d}_{i}\left(f_{i}\left(x\right),f_{i}\left(x_{0}\right)\right)<\varepsilon\right\} \right\} \\ = & \left\{ x\in\R\vert\sup_{i\in\N}\left|nx-nx_{0}\right|<\varepsilon\right\} \\ = & \left\{ x\in\R\vert\left|x-x_{0}\right|<\frac{\varepsilon}{n}\mbox{ for all \ensuremath{n\in\N} }\right\} \\ = & \left\{ x\in\R\vert\left|x-x_{0}\right|\leq0\right\} \\ = & \left\{ x_{0}\right\} . \end{align} Since $\left\{ x_{0}\right\}$ is closed and $B_{\bar{\rho}}\left(f\left(x_{0}\right),\varepsilon\right)$ is open, $f\left(t\right)$ is not continuous.

(b).

We also divide into several cases.

In the $\textbf{product topology}$, since $\left(\pi_{i}\left(\mathbf{x}_{n}\right)\right)_{n=1}^{\infty}\rightarrow0$ for all$i\in\N,$ in view of Exercise 19.6 we conclude that $\mathbf{x}_{n}\rightarrow\mathbf{0}.$ Similarly, we have $\mathbf{w}_{n}\rightarrow\mathbf{0}$, $\mathbf{y}_{n}\rightarrow\mathbf{0}$ and $\mathbf{z}_{n}\rightarrow\mathbf{0}$ as well.

In the $\textbf{box topology}$, we only have $\mathbf{z}_{n}\rightarrow\mathbf{0}$. First, we note that \begin{align} \mbox{product topology}\subset & \mbox{uniform topology}\subset\mbox{box topology}; \end{align} hence if any of these sequences converges in either uniform or box topology, it would converges to the same limit in product topology, namely $\mathbf{0}.$

We first show that $\mathbf{z}_{n}\rightarrow\mathbf{0}.$ Let $U=\prod_{n=1}^{\infty}U_{n}$ be an open neighborhood of $\mathbf{0}$ in the box topology. Since $0\in U_{1}$ and $0\in U_{2}$ and $U_{1},U_{2}$ are open in $\R$, we can find $\delta_{1},\delta_{2}$ such that $0\in\left(-\delta_{1},\delta_{1}\right)\subset U_{1}$ and $0\in\left(-\delta_{2},\delta_{2}\right)\subset U_{2}$. For all $n\geq\max\left\{ \lfloor1/\delta_{1}\rfloor+1,\lfloor1/\delta_{2}\rfloor+1\right\}$, we have \begin{align} \mathbf{z}_{n}\in & \left(-\delta_{1},\delta_{1}\right)\times\left(-\delta_{2},\delta_{2}\right)\times U_{3}\times\cdots\times U_{n}\times\cdots\\ \subset & \prod_{n=1}^{\infty}U_{n}; \end{align} hence $\mathbf{z}_{n}\rightarrow\mathbf{0}.$

Next, we show that none of $\left\{ \mathbf{x}_{n}\right\}$, $\left\{ \mathbf{w}_{n}\right\}$ and $\left\{ \mathbf{y}_{n}\right\}$ converges. Let $B=\prod_{n=1}^{\infty}B_{n}$ where $B_{n}=\left(-1/n,1/n\right)$. Since $\mathbf{0}\in B$, if any of $\left\{ \mathbf{w}_{n}\right\}$, $\left\{ \mathbf{x}_{n}\right\}$ any $\left\{ \mathbf{y}_{n}\right\}$ converges, $B$ will contain infinitely many terms of it. Clearly, this is not the case for $\mathbf{w}_{n}$; as for any $n\in\N$, we have $\pi_{n}\left(\mathbf{w}_{n}\right)=n\notin B_{n}$; as a result $\mathbf{w}_{n}\notin\prod_{n=1}^{\infty}B_{n}=B$ for all $n\in\N.$ Secondarily, we note that for any $n\in\N,$ $\pi_{i}\left(\mathbf{x}_{n}\right)=1/n$ for all $i\in\N$ and $i\geq n$. Since $1/n\notin B_{n}$, $\mathbf{x}_{n}\notin B$ for all $n\in\N.$ Similarly, since $\pi_{n}\left(\mathbf{y}_{n}\right)=1/n\notin B_{n}$, we have $\mathbf{y}_{n}\notin B$ for all $n\in\N$ as well. Thus, none of $\left\{ \mathbf{x}_{n}\right\}$, $\left\{ \mathbf{y}_{n}\right\}$ and $\left\{ \mathbf{w}_{n}\right\}$ converges in the box topology.

In the $\textbf{uniform}$ $\textbf{topology}$, we have $\mathbf{x}_{n}\rightarrow\mathbf{0}$ and $\mathbf{z}_{n}\rightarrow0$ and $\mathbf{y}_{n}\rightarrow0$ and $\mathbf{w}_{n}$ doesn't converge. First, since $\mathbf{z}_{n}$ converges in the box topology, which is finer than the uniform topology, $\mathbf{z}_{n}\rightarrow0$ }as desired. Secondarily, since for any $n\in\N$ we have \begin{align} \bar{\rho}\left(\mathbf{x}_{n},0\right)= & \sup_{i\in\N}\left\{ \bar{d}\left(x_{i},0\right)\right\} =\frac{1}{n}, \end{align} for any $\varepsilon>0$ we have $\mathbf{x}_{n}\in B_{\bar{\rho}}\left(\mathbf{0},\varepsilon\right)$ for all $n\geq\lfloor1/\varepsilon\rfloor+1$. Hence $\mathbf{x}_{n}\rightarrow\mathbf{0}$ since any open set $U$ containing $\mathbf{0}$ in the uniform topology would contain an open ball $B_{\bar{\rho}}\left(\mathbf{0},\delta\right)$ for some $\delta>0$ and $B_{\bar{\rho}}\left(\mathbf{0},\delta\right)\subset U$. In the same fashion, we can show that $\mathbf{y}_{n}\rightarrow\mathbf{0}$ as well. Finally, we note that since for any $n\in\N$ we have \begin{align} \bar{\rho}\left(\mathbf{0},\mathbf{w}_{n}\right)= & \sup_{i\in\N}\left\{ \bar{d}\left(w_{i},0\right)\right\} =1, \end{align} as a result, $\mathbf{w}_{n}\notin B_{\bar{\rho}}\left(\mathbf{0},\varepsilon\right)$ for any $\varepsilon\in\left(0,1\right)$; hence $\mathbf{w}_{n}$ doesn't converge.

• looks pretty good to me. – DanielWainfleet Nov 15 '15 at 16:31