# 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?

• I think your approach is more or less correct, but be wary of the details, I learned topology from this book and recall that a lot of the harder problems have some insidious subtleties to them that are easy to miss. For this problem, the one that's nagging me looking at your argument is whether you've really shown that the inverse image of any open set $V$, $h^{-1}(V)$ is actually equal to an open set, $U$, and not just contained in one. – Justin Benfield Nov 2 '16 at 18:30
• @JustinBenfield. $h$ is a bijection (although the OP did not say so).But he has shown that $h$ is continuous with respect to the uniform topology, and the similar argument shows that $h^{-1}$ is continuous. All of this suffices to conclude that $h$ is a homeomorphism . – DanielWainfleet Nov 3 '16 at 5:12
• @user254665 is my solution correct? And, what are the necessary conditions on the $a_n$ if $h$ is continuous? if $h$ is a homeomorphism? – Saaqib Mahmood Nov 3 '16 at 12:22
• @user254665 I am aware that a homeomorphism is a continuous bijection whose inverse is also continuous. I'm not taking any issue with that part, rather it's with his proof that $h$ (or $h^{-1}$) is, in fact, continuous that I am expressing skepticism of. – Justin Benfield Nov 3 '16 at 15:27
• @JustinBenfield "So if $|a_n|\leq 1$ for all $n$" in the proof that h is coninuous: But what if the sequence $(a_n)_n$ is unbounded? There seems to be a flaw. – DanielWainfleet Nov 3 '16 at 17:51

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_it_i-\delta $$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.