This is homework question. I just need hint. Please do not post solution.

Given sequence $(a_1, a_2,\ldots), (b_1,b_2,\ldots) \in \mathbb{R}^{\omega}$(countable cartesian product of real numbers), and that $a_i >0 \quad \forall i$. Define $h : \mathbb{R}^{\omega} \to \mathbb{R}^{\omega}$ by

$h((x_1,x_2,\ldots))=(a_1 x_1 +b_1, a_2 x_2+b_2,\ldots)$

Show that if $\mathbb{R}^{\omega}$ is given the product topology, then $h$ is a homeomorphism of $\mathbb{R}^{\omega}$ to $\mathbb{R}^{\omega}$. What happens if $\mathbb{R}^{\omega}$ is given the box topology?

Attempt: Goal is to show that $h$ and $h^{-1}$ is continuous. Let $U$ be an open set in product topology ($U$ is of the form $\prod_n X_n$, where $X_n = \mathbb{R}$ except for finitely many $n$), and now take $h^{-1}(U)$. I think it is wrong aproach because I do not think that $h^{-1}(U)$ is even defined in this case hence I could not really talk about whether the preimage is continuous or not.

Also I tried to use coordinate projection but I do not get really far away with this.

  • 1
    $\begingroup$ If you know that $\mathbb{R}^\omega$ with its product topology is first countable, you can verify continuity by using convergent sequences. And there's a pretty simple description of the convergent sequences in $\mathbb{R}^\omega$... $\endgroup$ – Nate Eldredge Feb 21 '13 at 2:16
  1. Not all open sets of the product space are of the given form, these just form a base for the topology. However, it is indeed enough to check that their inverse images are open.
  2. $h^{-1}(U)$ is a more general notation, and is defined for any function $h$ (even if it is not injective), and is called the preimage of the set $U$, and is defined as $$h^{-1}(U)=\{x\mid h(x)\in U\}\ .$$
  3. $h$ is a composition of functions $(x_n)_n\mapsto (a_nx_n)_n$ and $(y_n)_n\mapsto (y_n+b_n)$, and in fact, also these are homeomorphisms, and you can easily find their inverse. Use that $a_n\ne 0$.
  4. Since $h^{-1}$ will look very similarly, it is enough to prove that $h(U)$ is open for a base open set $U$.

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