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