Recall that $h((x_{1},x_{2},\ldots))=(a_{1}x_{1}+b_{1},a_{2}x_{2}+b_{2},\ldots)$ , $a_{i}>0$ , $h$ is bijective, and $h^{-1}(\mathbf{y})=$ $(\tfrac{1}{a_{1}}y_{1}-\tfrac{b_{1}}{a_{1}},\tfrac{1}{a_{2}}y_{2}-\tfrac{b_{2}}{a_{2}},\ldots)$ . So, if we find the conditions on $a_{i}$ ’s and $b_{i}$ ’s such that $h$ is continuous, the same conditions on $\tfrac{1}{a_{i}}$ ’s and $\tfrac{-b_{i}}{a_{i}}$ ’s will ensure that $h^{-1}$ is continuous. The combination of these two sets of conditions will give us the conditions under which $h$ is a homeomorphism.

According to the solution of Exercise 6(a), $A=h^{-1}(B_{\overline{\rho}}(\mathbf{y},\epsilon))\subset$ $h^{-1}(U(\mathbf{y},\epsilon))=$ $\prod_{n\in\mathbb{Z}_{+}}(\tfrac{1}{a_{n}}y_{n}-\tfrac{b_{n}}{a_{n}}-\tfrac{\epsilon}{a_{n}},\tfrac{1}{a_{n}}y_{n}-\tfrac{b_{n}}{a_{n}}+\tfrac{\epsilon}{a_{n}})=$ $\prod_{n\in\mathbb{Z}_{+}}A_{n}$ . Note, that if $\{a_{n}\}$ is unbounded, then for any $\delta>0$ , there is some $n$ such that $\tfrac{\epsilon}{a_{n}}<\delta$ , and the diameter of the interval $A_{n}$ is less than $2\delta$ , so that $A$ cannot contain any ball of size $\delta$ . Hence, one condition on $a_{i}$ ’s is that $\{a_{i}\}$ must be bounded. Now, if for all $n\in\mathbb{Z}_{+}$ , $a_{n}\le M$ , then for every $\mathbf{x}\in A$ , according to Exercise 6(c), there is $\delta<\epsilon$ such that $\mathbf{x}\in h^{-1}(U(\mathbf{y},\delta))$ , and for every $n\in\mathbb{Z}_{+}$ , $x_{n}\in(\tfrac{1}{a_{n}}y_{n}-\tfrac{b_{n}}{a_{n}}-\tfrac{\delta}{a_{n}},\tfrac{1}{a_{n}}y_{n}-\tfrac{b_{n}}{a_{n}}+\tfrac{\delta}{a_{n}})$ . Hence, if $\mathbf{z}\in B_{\overline{\rho}}(\mathbf{x},\tfrac{\epsilon-\delta}{2M})\subset U(\mathbf{x},\tfrac{\epsilon-\delta}{2M})$ , then $z_{n}\in(x_{n}-\tfrac{\epsilon-\delta}{2M},x_{n}+\tfrac{\epsilon-\delta}{2M})\subset$ $(\tfrac{1}{a_{n}}y_{n}-\tfrac{b_{n}}{a_{n}}-(\tfrac{\epsilon-\delta}{2M}+\tfrac{\delta}{a_{n}}),\tfrac{1}{a_{n}}y_{n}-\tfrac{b_{n}}{a_{n}}+(\tfrac{\epsilon-\delta}{2M}+\tfrac{\delta}{a_{n}}))$ where $a_{n}(\tfrac{\epsilon-\delta}{2M}+\tfrac{\delta}{a_{n}})=$ $a_{n}\tfrac{\epsilon-\delta}{2M}+\delta\le$ $\tfrac{\epsilon-\delta}{2}+\delta=\tfrac{\epsilon+\delta}{2}$ , and $h(\mathbf{z})\in U(\mathbf{y},\tfrac{\epsilon+\delta}{2})\subset B_{\overline{\rho}}(\mathbf{y},\epsilon)$ .

To summarize, $h$ is continuous in the uniform topology iff $\{a_{i}\}$ is bounded, i.e. there is some $M$ such that $a_{i}\le M$ for all $i\in\mathbb{Z}_{+}$ . Similarly, $h^{-1}$ is continuous iff $\tfrac{1}{a_{i}}$ is bounded, or, in other words, iff there is some $m>0$ such that $a_{i}\ge m$ for all $i\in\mathbb{Z}_{+}$ . Hence, $h$ is a homeomorphism iff there are $m,M\in\mathbb{R}_{+}$ such that for every $i\in\mathbb{R}_{+}$ , $m\le a_{i}\le M$ .