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

*

*What are some more "practical" examples of spaces with the condition above?

*What are other spaces or conditions for spaces being homeomorphic to $\mathbb{Q}^\omega$?

Thank you for any advice.
 A: 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).

*

*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 $X \simeq \mathbb{Q}^\mathbb{N}$.



*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 $A$, $\{ f \in H(X) : f(A) = A\}$.



*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).



*Paper: Countable products of zero-dimensional absolute Fσδ spaces

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



*Paper: The same as 5)


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



*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 $\mathbb{Q}^\omega$.

A: 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.
