Exercise 19.8

Proof. Consider any basis element $B=\prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }$ in $\prod <!--\; FUNCTION\; APPLICATION\; -->Y_{\alpha }$ so that each $V_{\alpha }$ is open in $Y_{\alpha }$ since we are in the box topology. For each $\alpha \in J$ then define $U_{\alpha }={f_{\alpha }}^{-1}(V_{\alpha })$, which is open in $X_{\alpha }$ since $f_{\alpha }$ is continuous. Hence the set $U=\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$ is a basis element of $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$ in the box topology and is therefore open. We claim that $U=f^{-1}(B)$, which shows that $f$ is continuous since $U$ is open and $B$ was arbitrary.

$\text{(⊂)}$ Consider any $x\in U=\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$. Then, for any $\alpha \in J$, we have $x_{\alpha }\in U_{\alpha }={f_{\alpha }}^{-1}(V_{\alpha })$ so that $f(x_{\alpha })\in V_{\alpha }$. Hence $f(x)={(f_{\alpha }(x_{\alpha }))}_{\alpha \in J}\in \prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }=B$ so that $x\in f^{-1}(B)$. this shows that $U\subset f^{-1}(B)$ since $x$ was arbitrary.

$\text{(⊃)}$ Now consider any $x\in f^{-1}(B)$ so that $f(x)\in B=\prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }$ and hence each $f_{\alpha }(x_{\alpha })\in V_{\alpha }$ by the definition of $f$. Then $x_{\alpha }\in {f_{\alpha }}^{-1}(V_{\alpha })=U_{\alpha }$ so that clearly $x\in \prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }=U$. Since $x$ was arbitrary this shows that $f^{-1}(B)\subset U$ as well. □

Proof. First note that clearly $h(x)=(h_1(x),h_2(x),\dots )$ for $x\in {ℝ}^{\omega }$, where each $h_i:{ℝ}^{\omega }\to ℝ$ is defined by

This can further be broken down as $h_i(x)=f_i({\pi }_i(x))=(f_i\circ {\pi }_i)(x)$, where each $f_i:ℝ\to ℝ$ is defined by $f_i(x)=a_{i}x+b_i$. As discussed in the proof of Theorem 19.6, each ${\pi }_i$ is continuous and we have that each $f_i$ is continuous by elementary calculus, noting that this is true whether each $a_i>0$ or not. It then follows from Theorem 18.2 part (c) that each $f_i\circ {\pi }_i=h_i$ is continuous. Then we have that $h$ is continuous by Theorem 19.6 since each coordinate function is continuous and we are using the product topology.

Now define the functions $g_i:ℝ\to ℝ$ by $g_i(x)=(x-b_i)∕a_i$ for $i\in {\mathbb{Z}}_{\text{+}}$, noting that this is defined since each $a_i>0$. Define also the functions $k_i:{ℝ}^{\omega }\to ℝ$ by $k_i=g_i\circ {\pi }_i$, and finally define $k:{ℝ}^{\omega }\to {ℝ}^{\omega }$ by $k(x)=(k_1(x),k_2(x),\dots )$. Now again we have that each ${\pi }_i$ and $g_i$ are continuous by the proof of Theorem 19.6 and elementary calculus. Hence $k_i=g_i\circ {\pi }_i$ and $k$ are continuous by Theorem 18.2 part (c), and Theorem 19.6, respectively, as before.

Now consider any $x=\left(x_1,x_2,\dots \right)\in {ℝ}^{\omega }$ so that we have, for any $i\in {\mathbb{Z}}_{\text{+}}$,

Therefore

We also have that

for each $i\in {\mathbb{Z}}_{\text{+}}$ so that

Since $x$ was arbitrary, it thus follows from Lemma 2.1 that $h$ is bijective and $k=h^{-1}$. Since we have already shown that $h$ and $k=h^{-1}$ are continuous, this suffices to prove that $h$ is a homeomorphism as desired. □

Proof. First, $h$ is still a bijection as the proof of this above does not depend on the topology at all. However, Theorem 19.6 was used in the proofs that $h$ and $h^{-1}$ are continuous, and we know that this theorem is not generally true for the box topology. On the other hand $h$ can be formulated as $h(x)=(f_1(x_1),f_2(x_2),\dots )$, where as before each $f_i(x)=a_{i}x+b_i$. Since each $f_i$ is continuous by elementary calculus, it follows from Lemma 1 that $h$ is continuous in the box topology. The same argument applies to the inverse function $h^{-1}$ since $h^{-1}(x)=(g_1(x_1),g_2(x_2),\dots )$ and each $g_i$ is continuous. □