What spaces are homeomorphic to $\mathbb{Q}^\omega$ = $\mathbb{Q}^\mathbb{N}$ = $\mathbb{Q}^\infty$?

Question

What spaces are homeomorphic to $\mathbb{Q}^\omega$ = $\mathbb{Q}^\mathbb{N}$ = $\mathbb{Q}^\infty$? (The space of all rational sequences, considered with the standard product topology).

I have found interesting characterization in a paper by Engelen "Characterizations of the countable infinite product of rationals and some related problems" that says:

Let $X = \{ (x_i)_{i \in \mathbb{N}} \in \mathbb{N}^\omega: lim_{i \mapsto \infty} x_i = \infty \}$. Then $X \simeq \mathbb{Q}^\omega$.

I find this interesting and my questions are

1. What are some more "practical" examples of spaces with the condition above?
2. What are other spaces or conditions for spaces being homeomorphic to $\mathbb{Q}^\omega$?

Thank you for any advice.

Answer 1

I have so far discovered four types of spaces homeomorphic to $\mathbb{Q}^\mathbb{N}$. (Thanks for useful sources that @Dave L. Renfo provided in his comments).

1. Paper: Characterizations of the countable infinite product of rationals and some related problems, author: Engelen (sorry I didnt find an online link, I have access via my university).

Let $X = \{ (x_i)_{i \in \mathbb{N}} \in \mathbb{N}^\omega: lim_{i \mapsto \infty} x_i = \infty \}$. Then $X \simeq \mathbb{Q}^\mathbb{N}$.

2. Paper: On the Group of Homeomorphisms of the Real Line That Map the Pseudoboundary Onto Itself

If $A$ is any countable dense subset of $\mathbb{R}$ then $H(\mathbb{R} | A)$ is homeomorphic to $\mathbb{Q}^\mathbb{N}$. (Where $H(\mathbb{R} | A)$ denotes the subgroup of $A$, $\{ f \in H(X) : f(A) = A\}$.

3. From the same paper as 2):

$H(C | D)$ is homeomorphic to $\mathbb{Q}^\mathbb{N}$. (Where $C$ stands for the Cantor set and $D$ is its countable dense subgroup).

4. Paper: Countable products of zero-dimensional absolute Fσδ spaces

For every zero-dimensional $F_{\sigma \delta}$-space $X$, we have $X \times \mathbb{Q}^\mathbb{N} \simeq \mathbb{Q}^\mathbb{N}$.

5. Paper: The same as 5)

Let $X$ be a non-empty closed subset of $\mathbb{Q}^\omega$. Then $X \times \mathbb{Q}^\omega \simeq \mathbb{Q}^\omega$.

6. The same paper as 4), 5)

Let $\{X_i: i \in \mathbb{N}\}$, $\{Y_i: i \in \mathbb{N}\}$ be families of non-empty, zero-dimensional absolute $F_{\sigma \delta}$-spaces which are not complete, and suppose that $\prod_{i=1}^\infty X_i$ and $\prod_{i=1}^\infty Y_i$ are homogeneous. If $\prod_{i=1}^\infty X_i$ and $\prod_{i=1}^\infty W_i$ are not Baire, then they are homeomorphic to $\mathbb{Q}^\omega$.

Answer 2

Let $f(z)=\exp(z)-1$. The "zero-dimensional remainder" of $\mathbb Q ^\omega$ is homeomorphic to the set of points in the complex plane $z\in \mathbb C$ such that $f^n(z)$ goes to neither $0$ nor $\infty$ https://arxiv.org/pdf/2010.13876.pdf. By zero-dimensional remainder, I mean the complement of a dense copy of $\mathbb Q ^\omega$ in a complete zero-dimensional space. For example, $\mathbb P ^\omega\setminus (\mathbb Q+\pi)^\omega$. I am unaware of a copy of $\mathbb Q ^\omega$ that is generated very simply in this manner, in complex dynamics.