Prob. 7, Sec. 20 in Munkres' TOPOLOGY, 2nd ed: The coordinate-wise linear self-map of $\mathbb{R}^\omega$

Question

Here's Prob. 7, Sec. 20 in the book Topology by James R. Munkres, 2nd edition:

Consider the map $h \colon \mathbb{R}^\omega \to \mathbb{R}^\omega$ defined in Exercise 8 of Sec. 19; give $\mathbb{R}^\omega$ the uniform topology. Under what conditions on the numbers $a_i$ and $b_i$ is $h$ continuous? a homeomorphism?

Now here is Exercise 8 of Sec. 19:

Given sequences $\left( a_1, a_2, a_3, \ldots \right)$ and $\left( b_1, b_2, b_3, \ldots \right)$ of real numbers with $a_i > 0$ for all $i$, define $h \colon \mathbb{R}^\omega \to \mathbb{R}^\omega$ by the equation $$ h \left( \left( x_1, x_2, x_3, \ldots \right) \right) = \left( a_1 x_1 + b_1, a_2 x_2 + b_2, a_3 x_3 + b_3, \ldots \right).$$ Show that if $\mathbb{R}^\omega$ is given the product topology, $h$ is a homeomorphism of $\mathbb{R}^\omega$ with itself. What happens if $\mathbb{R}^\omega$ is given the box topology?

My effort:

For any $x, y \in \mathbb{R}^\omega$, we have $$ \tilde{\rho}(h(x), h(y) ) = \sup \left\{ \ \min \left\{ \ \left\vert a_n \right\vert \left\vert x_n - y_n \right\vert , 1 \right\} \ \colon \ n \in \mathbb{N} \ \right\}.$$ So if $\left\vert a_n \right\vert \leq 1$ for all $n \in \mathbb{N}$, then we obtain $$\tilde{\rho} ( h(x), h(y)) \leq \tilde{\rho}(x,y),$$ and so, given a real number $\varepsilon > 0$, if we take a real number $\delta$ such that $0 < \delta \leq \varepsilon$, then $$\tilde{\rho} ( h(x), h(y)) < \varepsilon$$ for all $x, y \in \mathbb{R}^\omega$ such that $$ \tilde{\rho}(x,y) < \delta.$$ Hence $h$ is uniformly continuous on $\mathbb{R}^\omega$. Am I right?

Now the inverse map $h^{-1} \colon \mathbb{R}^\omega \to \mathbb{R}^\omega$ is defined by $$ h^{-1}(x) = \left( \frac{x_1 - b_1}{a_1}, \frac{x_2 - b_2}{a_2}, \frac{x_3 - b_3}{a_3}, \ldots \right) $$ or $$ h^{-1}(x) = \left( \frac{1}{a_1} x_1 - \frac{b_1}{a_1}, \frac{1}{a_2} x_2 - \frac{b_2}{a_2}, \frac{1}{a_3} x_3 - \frac{b_3}{a_3}, \ldots \right) \ \ \ \mbox{ for all } \ x \colon= \left( x_1, x_2, x_3, \ldots \right) \in \mathbb{R}^\omega.$$ So, using what we have shown for $h$, we can conclude that, if $\left\vert a_n \right\vert \geq 1$ for all $n \in \mathbb{N}$, then $h^{-1}$ is uniformly continuous on $\mathbb{R}^\omega$. Am I right?

Therefore if $\left\vert a_n \right\vert = 1$ for all $n \in \mathbb{N}$, then $h$ is a homeomorphism. Am I right?

If what I've derived so far is correct, then does the converse of the above hold as well?

PS:

Here is my latest insight:

Suppose the sequence $\left( a_1, a_2, a_3, \ldots \right)$ is unbounded. Then for any natural number $k$, there is a natural number $n_k$ such that $$ a_{n_k} > k. \tag{0} $$ And, for any point $$\mathbf{x} \colon= \left( x_n \right)_{n \in \mathbb{N} }, $$ in $\mathbb{R}^\omega$, and, for any real number $\delta > 0$, if we put $$\mathbf{y} \colon= \left( x_n + \frac{\delta}{2} \right)_{n \in \mathbb{N}}, \tag{1} $$ then $$ \begin{align} \bar{\rho} ( \mathbf{x}, \mathbf{y} ) &= \sup \big\{ \ \min \left\{ \ \lvert x_n - y_n \rvert, \ 1 \ \right\} \ \colon \ n \in \mathbb{N} \ \big\} \\ &\leq \sup \big\{ \ \lvert x_n - y_n \rvert \ \colon \ n \in \mathbb{N} \ \big\} \\ &= \frac{\delta}{2} \\ &< \delta. \end{align} \tag{2} $$ And, if $N$ is a natural number such that $N > 2/\delta$, then we have $$ N \frac{\delta}{2} > 1, \tag{3} $$ and we see that $$ \begin{align} \bar{\rho} \big( h \left( \mathbf{x} \right) , h \left( \mathbf{y} \right) \big) &= \bar{\rho} \big( \left( a_n x_n + b_n \right)_{n \in \mathbb{N} }, \left( a_n x_n + b_n \right)_{n \in \mathbb{N} } \big) \\ &= \sup \big\{ \ \min \left\{ \ \left\lvert \left( a_n x_n + b_n \right) - \left( a_n y_n + b_n \right) \right\rvert, \ 1 \ \right\} \ \colon \ n \in \mathbb{N} \ \big\} \\ &= \sup \big\{ \ \min \left\{ \ \left\lvert a_n \right\rvert \left\lvert x_n - y_n \right\rvert , \ 1 \ \right\} \ \colon \ n \in \mathbb{N} \ \big\} \\ &= \sup \big\{ \ \min \left\{ \ a_n \left\lvert x_n - y_n \right\rvert , \ 1 \ \right\} \ \colon \ n \in \mathbb{N} \ \big\} \qquad \mbox{ [ because $a_n > 0$ for all $n$ ] } \\ &\geq \sup \big\{ \ \min \left\{ \ a_{n_k} \left\lvert x_{n_k} - y_{n_k} \right\rvert , \ 1 \ \right\} \ \colon \ k \in \mathbb{N} \ \big\} \\ &\geq \sup \big\{ \ \min \left\{ \ k \left\lvert x_{n_k} - y_{n_k} \right\rvert , \ 1 \ \right\} \ \colon \ k \in \mathbb{N} \ \big\} \qquad \mbox{ [ using (0) above ] } \\ &= \sup \left\{ \ \min \left\{ \ k \frac{\delta}{2} , \ 1 \ \right\} \ \colon \ k \in \mathbb{N} \ \right\} \qquad \mbox{ [ using (1) above ] } \\ &\geq \min \left\{ \ N \frac{\delta}{2} , \ 1 \ \right\} \\ &= 1 \qquad \mbox{ [ using (3) above ] } \\ &> \varepsilon \end{align} \tag{4} $$ whenever $\varepsilon$ is any real number such that $0 < \varepsilon < 1$.

Thus we have shown that if we take $\varepsilon \in (0, 1)$, then, for any real number $\delta > 0$, there is a point $\mathbf{y} \in \mathbb{R}^\omega$ such that $$ \bar{\rho} ( \mathbf{x}, \mathbf{y} ) < \delta, $$ but $$ \bar{\rho} \big( h \left( \mathbf{x} \right) , h \left( \mathbf{y} \right) \big) > \varepsilon. $$

Thus if the sequence $\left( a_n \right)_{n \in \mathbb{N} }$ is unbounded (from above), then the function $h$ cannot be continuous at any point $\mathbf{x}$ of $\mathbb{R}^\omega$.

So let us assume that the sequence $\left( a_n \right)_{n \in \mathbb{N} } $ is bounded (above). Then there is a positive real number $M$ such that $a_n < M$ for all $n$.

So, for any given real number $\varepsilon > 0$, if we take any real number $\delta$ such that $$ 0 < \delta < \min\left\{ \ \frac{\varepsilon}{2M}, \ 1 \ \right\}, $$ then for any points $\mathbf{x}$, $\mathbf{y}$ in $\mathbb{R}^\omega$ which satisfy $$ \bar{\rho}( \mathbf{x}, \mathbf{y} ) < \delta, $$ we see that for each $n \in \mathbb{N}$, we have the inequality $$ \min \left\{ \ \left\lvert x_n - y_n \right\rvert, \ 1 \ \right\} \leq \bar{\rho}(\mathbf{x}, \mathbf{y} ) < \delta < 1, $$ and so $$ \min \left\{ \ \left\lvert x_n - y_n \right\rvert, \ 1 \ \right\} = \left\lvert x_n - y_n \right\rvert, $$ and hence $$ \left\lvert x_n - y_n \right\rvert < \frac{\varepsilon}{2M}. $$ Therefore, $$ \begin{align} \bar{\rho}\left( h(\mathbf{x}), h(\mathbf{y}) \right) &= \sup \left\{ \ \min \left\{ \ a_n \left\lvert x_n - y_n \right\rvert, \ 1 \ \right\} \ \colon \ n \in \mathbb{N} \ \right\} \\ &\leq \sup \left\{ \ \min \left\{ \ M \left\lvert x_n - y_n \right\rvert, \ 1 \ \right\} \ \colon \ n \in \mathbb{N} \ \right\} \\ &\leq M \sup \left\{ \ \min \left\{ \ \left\lvert x_n - y_n \right\rvert, \ 1 \ \right\} \ \colon \ n \in \mathbb{N} \ \right\} \\ &= M \sup \left\{ \ \left\lvert x_n - y_n \right\rvert \ \colon \ n \in \mathbb{N} \ \right\} \\ &\leq M \frac{\varepsilon}{2M} \\ &= \frac{\varepsilon}{2} \\ &< \varepsilon. \end{align} $$ Hence $h$ is (uniformly) continuous (on all of $\mathbb{R}^\omega$).

Is this proof correct. If so, then is each and every step in it correct and clear enough? If not, then where lies the problem?

Answer 1

Nice proof . Here's an alternative proof using exercise 20.6.c:

Consider an arbitrary image point $h(\vec{t})$ and a basic set $U= \bigcup\limits_{\delta<\epsilon} \prod\limits_{i=1}^{\infty}(a_it_i+b_i-\delta, a_it_i+b_i+\delta)$ about that image point. Then any other image point $h(\vec{s})\in U$ has the form $h(\vec{s})=(a_1s_1+b_1, a_2s_2+b_2,...)$ where in each component $a_is_i+b_i \in (a_it_i+b_i-\delta, a_it_i+b_i+\delta)$. That is, for all $\delta<\epsilon$:

$$a_it_i+b_i-\delta<a_is_i+b_i<a_it_i+b_i+\delta$$ $$a_it_i-\delta<a_is_i<a_it_i+\delta$$ $$t_i-\frac{\delta}{a_i}<s_i<t_i+\frac{\delta}{a_i}$$

The union of all such $\vec{s}$ and $\delta$ is $h^{-1}(U)=\bigcup\limits_{\delta<\epsilon} \prod\limits_{i=1}^{\infty}(t_i-\frac{\delta}{a_i}, t_i+\frac{\delta}{a_i})$. It's reasonable to conjecture that this is an open set if and only if $(a_i)_{i \in \mathbb{N}}$ is bounded. For $a_i$ is not bounded, I think $f(\vec{x})=(x_1, 2x_2, 3x_3,...)$ is a counterexample, since the preimage of a basic open set about $f(\vec{0})$ is ${0}$, which is not closed. To show the contrapositive, we must show that if $(a_i)_{i \in \mathbb{N}}$ is bounded, no matter how small an $\epsilon$ you choose, $\exists \beta$ such that if $x\in B_{\bar{\rho}}(\vec{s},\beta)$ then $f(\vec{x})\in B_{\bar{\rho}}(f(\vec{s}), \epsilon)$. That is to say: every preimage point has an entire basis element in the preimage, meaning the preimage is an infinite union of bases, meaning the preimage is open.

If $x\in B_{\bar{\rho}}(\vec{s},\beta)=B_{\bar{\rho}}(\vec{s},\frac{\epsilon}{4\max((a_i)_{i\in \mathbb{N}})})$ (that choice is motivated later on) then for some $j<\beta$, $x_i\in (s_i-j,s_i+j) ,\forall i$, as a result of of 6c. So certainly $x_i\in (s_i-\beta,s_i+\beta) ,\forall i$. Then taking the extreme values of $s_i$, we see that $x_i\in(t_i-\frac{\delta}{a_i}-\beta,t_i-\frac{\delta}{a_i}+\beta)$. Then by plugging in extreme values of $x_i$ to $h$, we'll get a range for $\pi_i(h(\vec{x}))$. We already have a range for $h(\vec{s})$. Then comparing the two, we see that the largest value for the difference is $2\beta a_i$. If we let, $\beta=\frac{\epsilon}{4\max((a_i)_{i\in \mathbb{N}})}$ then we make $\bar{\rho}(\pi_i(f(\vec{x})-f(\vec{s})))<2\beta a_i<\epsilon/2<\epsilon, \forall i$. The reason I "overkilled" the $\beta$ choice was to ensure that the supremum doesn't somehow become $\epsilon$, although my hunch is that wasn't necessary. So $h(\vec{x})\in U(\vec{s},\frac{\epsilon}{2})\subset B_{\bar{\rho}}(f(\vec{s}), \epsilon)$.

This establishes the continuity of $h$. By simple algebra, $\pi_i(h^{-1}(\vec{x}))=\frac{x_i}{a_i}-\frac{b_i}{a_i}$. Note that this has the same format as $h$. Hence a similar proof will show that $h^{-1}$ is continuous if and only if $(\frac{1}{a_i})_{i\in \mathbb{N}}$ is bounded.

To recap: we used 6c to describe the preimage of open sets in the range, with a funky formula. We noticed immediately that $b_i$ was irrelevant. We then conjectured that the funky formula would be open iff $a_i$ was bounded. We gave a counterexample when $a_i$ wasn't bounded, then showed that if $a_i$ is bounded, the preimage of any basic open set is open. We did this by showing every point in the preimage has a basic set containing that point and also contained in the preimage. I'm taking topology now by the way, so let me know if I screwed up somewhere.

I wonder if anyone can prove this using Theorem 18.1(2) and the result of exercise 20.6.c. A general idea would be that the closure of a union is the union of a closure, and the closure of a product is the product of the closures. I didn't pursue that idea much myself, but I'd be interested to see what someone comes up with.