Given sequences \((a_1, a_2, \ldots)\) and \((b_1, b_2, \ldots)\) of real numbers with \(a_i > 0\) for all \(i\), define \(h: \mathbb{R}^\omega \to \mathbb{R}^\omega\) by the equation
\[ h((x_1, x_2, \dots)) = (a_1 x_1 + b_1, a_2 x_2 + b_2, \dots). \]
Show that if \(\mathbb{R}^\omega\) is given the product topology, \(h\) is a homeomorphism of \(\mathbb{R}^\omega\) with itself. What happens if \(\mathbb{R}^\omega\) is given the box topology?
Solution
For each \(i\), let \[ h_i:\mathbb R\to\mathbb R,\qquad h_i(x)=a_ix+b_i. \] Since \(a_i>0\), \(h_i\) is a homeomorphism with inverse \[ h_i^{-1}(y)=\frac{y-b_i}{a_i}. \] The map \(h\) is the product map \(\prod_i h_i\).
For the product topology, the inverse image of a basic set \[ \prod_iU_i, \] with \(U_i=\mathbb R\) except for finitely many \(i\), is \[ \prod_i h_i^{-1}(U_i), \] which is again basic product-open. The same statement holds for \(h^{-1}\), so \(h\) is a homeomorphism.
Exactly the same proof works for the box topology: now every coordinate may be restricted, but \[ h^{-1}\left(\prod_iU_i\right)=\prod_i h_i^{-1}(U_i) \] is a basic box-open set, and likewise for the inverse map. Thus \(h\) is a homeomorphism for both the product and box topologies.