Does $h$ exhibits a homeomorphism between $\mathbb R^{J}$ and itself in the Box Topology?

Question

Does the proof here (Does the given $h$ exhibit the homeomorphism between $\mathbb{R}^{\omega}$ and itself with box topologies?) works when we are dealing with $\mathbb R^J$ instead of $\mathbb R^{\omega}$? I think that it should work as we are making for each $i \in \mathbb N$ a open. Is it okay for doing the same for each $\alpha \in J$? The proof in my mind is something like that:
Assume $h:\mathbb R^J \rightarrow \mathbb R^J$ that $h((x_{\alpha})_{\alpha \in J}):= ((a_{\alpha}.x_{\alpha}+b_{\alpha})_{\alpha \in J})$. Additionally, $(a_{\alpha})_{\alpha \in J}$ are positive real numbers and $(b_{\alpha})_{\alpha \in J}$ are real numbers. $$\forall \alpha \in J, \forall \; U=\prod_{\beta \in J}(x_\beta,y_\beta) \Rightarrow h^{-1}(U)=\prod_{\beta \in J}(\frac{x_\beta-b_\beta}{a_\beta},\frac{y_\beta-b_\beta}{a_\beta})$$

Answer 1

If $h_j, j \in J: \Bbb R \to \Bbb R$ is a homeomorphism, then $H: \Box \Bbb R^J \to \Box \Bbb R^J$ is a homeomorphism too, where $$H(x_j)_{j \in J}) = (h_j(x_j))_{j \in J}$$

That $H$ is a bijection is clear, and that $H$ is continuous follows from $$H^{-1}[\Box_{j \in J} U_j] = \Box_{j \in J} h_j^{-1}[U_j]$$ and the same holds for forward images

$$H[\Box_{j \in J} U_j] = \Box_{j \in J} h_j[U_j]$$ showing that $H$ is continuous and open.

Your map is a special case.