Does the given $h$ exhibit the homeomorphism between $\mathbb{R}^{\omega}$ and itself with box topologies?

Question

(Munkres, p. 118, Problem 8)

Give $h: \mathbb{R}^{\omega} \to \mathbb{R}^{\omega}$ $$ h(x_1,x_2,\dots) = (ax_1 +b,ax_2+b,\dots) $$ where $a,b\in \mathbb{R}$.

Does the given $h$ exhibit the homeomorphism between $\mathbb{R}^{\omega}$ and itself with box topologies?

My proof goes like this:

Since each $x_i \mapsto ax_i +b$ is continuous, for each open ball of $B_i(ax_i +b, \epsilon_i)$, we can find $B_i(x,\delta_i)$ such that $aB_i(x,\delta_i) +b \subseteq B_i(ax_i +b, \epsilon_i)$. Now construct $$ B_1(x_i,\delta_i) \times B_2(x_i,\delta_i) \times \cdots. $$ Then $$ h(B_1(x_i,\delta_i) \times B_2(x_i,\delta_i) \times \cdots) \subseteq B_1(ax_1 +b, \epsilon_1) \times B_2(ax_2 +b, \epsilon_2) \times \cdots $$ proving the continuity.

Would this be valid?

Answer 1

If $a \ne 0$, $h$ is invertible with $\def\R{\mathbb R}\def\Ro{\R^{\omega}}$ $$ h^{-1}\colon \Ro \to \Ro, \quad x \mapsto (a^{-1}x_1 - a^{-1}b, a^{-1}x_2 - a^{-1}b, \ldots) $$ so $h^{-1}$ is of "the same type" as $h$, so we will prove that $h$ is continuous, then $h^{-1}$ is continuous also (along the same lines).

So let $U = \prod_{i< \omega} (\alpha_i, \beta_i)$ a basis set of the box topology. We have if $a > 0$: \begin{align*} ax_i + b &\in (\alpha_i, \beta_i)\\ \iff ax_i &\in (\alpha_i - b, \beta_i - b)\\ \iff x_i &\in (a^{-1}\alpha_i - a^{-1}b, a^{-1}\beta_i - a^{-1}b) \end{align*} so $h^{-1}(U) = \prod_{i<\omega}(a^{-1}\alpha_i - a^{-1}b, a^{-1}\beta_i - a^{-1}b)$ is Box-open, hence $h$ is continuous, if $a < 0$, we have analogously \begin{align*} ax_i + b &\in (\alpha_i, \beta_i)\\ \iff ax_i &\in (\alpha_i - b, \beta_i - b)\\ \iff x_i &\in (a^{-1}\beta_i - a^{-1}b, a^{-1}\alpha_i - a^{-1}b) \end{align*} so $h^{-1}(U) = \prod_{i<\omega}(a^{-1}\beta_i - a^{-1}b, a^{-1}\alpha_i - a^{-1}b)$, which is also open.

Answer 2

HINT: The map $h$ is a composition of the maps

$$h_1:\Bbb R^\omega\to\Bbb R^\omega:\langle x_0,x_1,x_2,\ldots\rangle\mapsto\langle ax_0,ax_1,ax_2,\ldots\rangle$$

and

$$h_2:\Bbb R^\omega\to\Bbb R^\omega:\langle x_0,x_1,x_2,\ldots\rangle\mapsto\langle x_0+b,x_1+b,x_2+b,\ldots\rangle\;;$$

if you can show that $h_1$ and $h_2$ are both homeomorphisms, it will follow immediately that $h=h_2\circ h_1$ is a homeomorphism.

• It’s very easy to show that $h_2$ is a homeomorphism: it’s just a translation by $\langle b,b,b,\ldots\rangle$ that takes open boxes to open boxes.

• It’s also pretty easy to show that $h_1$ takes open boxes to open boxes and is therefore a homeomorphism unless $a$ has a certain value; what is that one exceptional value?

(It’s possible to work directly with $h$, but I think that it’s conceptually easier to work with the simpler maps $h_1$ and $h_2$.)