Clearly, $h$ is bijective (as a product of bijective functions). Also note, that the inverse function has the same form with different coefficients: $g(\mathbf{y})=(\tfrac{1}{a_{1}}y_{1}-\tfrac{b_{1}}{a_{1}},\tfrac{1}{a_{2}}y_{2}-\tfrac{b_{2}}{a_{2}},\ldots)$ . So, all we need to prove is that $h$ is continuous. Take a point $y=h(x)\in U$ where $U$ is open in the product, then $y$ has an open basis neighborhood $B\subset U$ , which is the product of some open sets $B_{n}\subset\mathbb{R}$ (and in the product topology only finitely many of $B_{n}$ ’s are proper subsets of $\mathbb{R}$ ). The preimage $A_{n}=\{x_{n}|a_{n}x_{n}+b_{n}\in B_{n}\}$ of each such set is open in $\mathbb{R}$ as well, and if $B_{n}=\mathbb{R}$ then $A_{n}=\mathbb{R}$ . Therefore, every point $x$ in $h^{-1}(U)$ has an open neighborhood $\prod_{n\in\mathbb{Z}_{+}}A_{n}\subset h^{-1}(U)$ (in the case of the product topology only finitely many of $A_{n}$ ’s are proper subsets of $\mathbb{R}$ ), and, therefore, $h^{-1}(U)$ is open, $h$ is continuous. This works for both product and box topologies.