# Prob. 8, Sec. 19, in Munkres' TOPOLOGY, 2nd ed: The map $\left( x_n \right) \mapsto \left( a_n x_n + b_n \right)$ is a homeomorphism

Here is Prob. 8, Sec. 19, in the book Topology by James R. Munkres, 2nd edition:

Given sequences $\left( a_1, a_2, \ldots \right)$ and $\left( b_1, b_2, \ldots \right)$ of real numbers with $a_i > 0$ for all $i$, define $h \colon \mathbb{R}^\omega \to \mathbb{R}^\omega$ by the equation $$h \big( \left( x_1, x_2, \ldots \right) \big) = \left( a_1 x_1 + b_1, a_2 x_2 + b_2, \ldots \right).$$ 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 Attempt:

Let $a$ and $b$ be real numbers such that $a > 0$, and let $f \colon \mathbb{R} \to \mathbb{R}$ be defined by $$f(t) = a t + b.$$ Then $f^\prime(t) = a> 0$ for all $t \in \mathbb{R}$ so that $f$ is a strictly increasing (and hence an injective) function throughout all of $\mathbb{R}$. Moreover, if $s \in \mathbb{R}$, then $$f \left( \frac{s-b}{a} \right) = a \frac{s-b}{a} + b = s-b + b = s.$$ Thus $f$ is bijective and $f^{-1} \colon \mathbb{R} \to \mathbb{R}$ is given by $$f^{-1} (s) = \frac{s - b}{a} = \frac{1}{a} s - \frac{b}{a}.$$

For the product topology:

From the preceding paragraph, we note that the function $h$ is bijective, and the inverse $h^{-1} \colon \mathbb{R}^\omega \to \mathbb{R}^\omega$ is given by the equation $$h^{-1} \big( \left( y_1, y_2, \ldots \right) \big) = \left( \frac{1}{a_1} y_1 - \frac{b_1}{a_1}, \frac{1}{a_2} y_2 - \frac{b_2}{a_2}, \ldots \right).$$

Let $$B \colon= \prod_{n =1 }^\infty B_n$$ be a basis element for the product topology on $\mathbb{R}^\omega$ such that for all $n$ in some finite set of natural numbers, say, $\left\{ n_1, \ldots, n_k \right\}$, we have $$B_n = \left( \alpha_n, \beta_n \right) = \left\{ \ t \in \mathbb{R} \ \colon \ \alpha_n < t < \beta_n \ \right\},$$ where $\alpha_n$ and $\beta_n$ are some real numbers such that $\alpha_n < \beta_n$, and such that $B_n = \mathbb{R}$ for all other $n$.

Then we note that $$h^{-1} \big( B \big) = \prod_{n=1}^\infty B_n^\prime,$$ where $$B_n^\prime = \begin{cases} \left( \frac{ \alpha_n - b_n}{a_n} , \frac{ \beta_n - b_n}{a_n} \right) \ & \mbox{ for } \ n = n_1, \ldots, n_k, \\ \mathbb{R} \ & \mbox{ otherwise}. \end{cases}$$ Thus $h^{-1}(B)$ is open in $\mathbb{R}^\omega$, showing that $h$ is continuous.

Moreover, $$h(B) = \prod_{n=1}^\infty B_n^*,$$ where $$B_n^* = \begin{cases} \left( \ a_n \alpha_n + b_n \ , \ a_n \beta_n + b_n \ \right) \ & \mbox{ for } \ n = n_1, \ldots, n_k, \\ \mathbb{R} \ & \mbox{ otherwise}. \end{cases}$$ Thus $h(B)$ is also open in $\mathbb{R}^\omega$, showing that $h^{-1}$ is also continuous. Hence $h$ is a homeomorphism.

Another way to prove this is as follows: As both the function $f$ and $f^{-1}$ in the very first paragraph of my solution are continuous, so each of the coordinate functions of both $h$ and $h^{-1}$ is continuous, and hence by Theorem 19.6 in Munkres both $h$ and $h^{-1}$ are continuous.

Now for the box topology:

Let $$B \colon= \prod_{n=1}^\infty \left( \alpha_n, \beta_n \right),$$ where for each natural number $n$, $\alpha_n$ and $\beta_n$ are real numbers such that $\alpha_n < b_n$. This $B$ is a box topology basis element for $\mathbb{R}^\omega$. Moreover, $$h(B) = \prod_{n=1}^\infty \left( a_n \alpha_n + b_n, \alpha_n \beta_n + b_n \right),$$ which is open in $\mathbb{R}^\omega$, thus showing that $h^{-1}$ is continuous, and also $$h^{-1}(B) = \prod_{n=1}^\infty \left( \frac{ \alpha_n - b_n }{ a_n }, \frac{ \beta_n - b_n }{a_n} \right),$$ which is again open in $\mathbb{R}^\omega$, thus showing that $h$ is continuous. Hence $h$ is again a homeomorphism.

Are my findings correct? If so, then is each step of my proof correct too? Is my presentation good enough?

• The proof and presentation both look fine to me. – pisco Sep 24 '17 at 6:36
• Try to be more concise. – Henno Brandsma Sep 24 '17 at 7:28
• @HennoBrandsma thank you for your advice; I'll try to comply with it. However, it is always my desire to write out my proof in sufficient detail so as to make these very easily understandable by a student who is at a very elementary in his studies. – Saaqib Mahmood Sep 24 '17 at 9:58

$f_n: X_n \to Y_n$ is continuous for all $n$, then $f:=\prod_n f_n$ defined on $X=\prod_n X_n$ to $Y=\prod_n Y_n$ (both in the product topology) defined by $(\prod_n f_n)((x_n)_n) = (f_n(x_n))_n$ is continuous too. This holds because for all $n$: $(\pi_Y)_n \circ (\prod_n f_n) = f_n \circ (\pi_X)_n$.
We apply this fact to the $h_n$ and their continuous inverses $h_n^{-1}$ (affine maps are continuous and the inverse of an affine map is affine).
The box topology is also clear as $(\prod_n f_n)^{-1}[\prod_n U_n] = \prod_n f_n^{-1}[U_n]$ and the same fact holds for box topologies as well by construction.