This question is motivated by this [MO-problem](https://mathoverflow.net/questions/302724/automorphisms-of-erd%C3%B6s-spaces/302839#302839) asking if the Erdős spaces $\mathfrak E$ and $\mathfrak E\_c$ admit a self-homeomorphism with dense orbits of points.


The affirmative answer would follow from the affirmative answer to the following 



> 
> **Problem.** Are the Erdős spaces $\mathfrak E$ and $\mathfrak E\_c$ homeomorphic to monothetic topological groups?
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Now I recall the necessary definitions. A topological group $G$ is *monothetic* if it contains a dense cyclic subgroup. Many examples of monothetic topological groups can be found [here](https://arxiv.org/pdf/math/9907128.pdf).


The subspaces
$$\mathfrak E:=\{(x\_n)\_{n\in\omega}\in\ell\_2:\forall n\in\omega\;\;x\_n\in\mathbb Q\}$$
and
$$\mathfrak E\_c:=\{(x\_n)\_{n\in\omega}\in\ell\_2:\forall n\in\omega\;\;x\_n\in\mathbb R\setminus\mathbb Q\}$$
of the real separable Hilbert space are called the *rational Erdős space* and the *complete Erdős space*, respectively. The rational Erdős space $\mathfrak E$ is a subgroup of $\ell\_2$ and the complete Erdős space $\mathfrak E\_c$ is homeomorphic to the closed subgroup $$G:=\{(x\_n)\_{n\in\omega}\in\ell\_2:\forall n\in\omega\;\; x\_n\in\tfrac1{2^n}\mathbb Z\}$$ by Theorem 4.11 of [Dijkstra and van Mill](https://pdfs.semanticscholar.org/629c/6bde87b1296d3b76a526ee99e31e32068770.pdf). 


However the subgroups $\mathfrak E$ and $G$ of $\ell\_2$ are not monothetic.



Has the Erdős space the structure of a monothetic topological group?
Let me try to supply some useful constructions, with a hope that they would help to solve the problems proposed by Taras (possibly, by an analogy rather than directly).



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> $\mathbf{\ell^2}$**-product of arbitrary metric spaces**
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The following construction must be classical. Let $\ (X\_n\ d\_n)\ $ be
metric spaces; let $\ p\_n\in X\_n;\ $ and let
$\ \mathbf X\_n:=((X\_n\ d\_n)\ p\_n).\ $ Define $\ m^2(\mathbf X)\ $ to be the set of all square summable sequences
$\ x:=(x\_1\ x\_2\ \ldots)\in \prod\_{n=1}^\infty X\_n,\ $ meaning that
$\ \sum\_{n=1}^\infty d\_n^2(x\_n\ p\_n) < \infty.\ $ The Euclid-Hilbert
distance $\ \delta\ $ in $\ m^2(\mathbf X)\ $ is given by


$$ \delta(x\ y):=\sqrt{\sum\_{n=1}^\infty d\_n^2(x\_n\ y\_n)}, $$


for $\ x:=(x\_1\ x\_2\ \ldots)\ $ and $\ y:=(y\_1\ y\_2\ \ldots),\ $
but of course.




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**Remark**   Let's consider the special case $\ \mathbf X\_1=\mathbf X\_2=\ldots.\ $
Then $\,\ m^2(\mathbf X)\,\ $ and
$\,\ m^2(\mathbf X)\times m^2(\mathbf X)\,\ $ are homeomorphic as topological spaces. Furthermore, we may introduce the Euclid-Hilbert
metrics in the latter space to make the two spaces **isometric**.



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> **An example of a monothetic group**
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Let $\ p\_1<p\_2\ldots\ $ be the sequence of all primes, and let
$\ \forall\_n\ q\_n=p\_n^2.\ $ Define a distance $\ d\_n\ $ in
$\ X\_n:= \mathbb Z\_{q\_n}:=\{0\ \ldots\ q\_n\!-\!1\}\ $ as follows:


$$ d\_n(s\ t)\ :=\ \frac 1{p\_n}\cdot\min(\,|s-t|\ \ \ q\_n\!-\!|s\!-\!t|\,) $$


Then we may introduce Euclid-Hilbert distance in the group
$\ \mathbf X\ :=\ \prod\_{n=1}^\infty X\_n.\ $ Element


$$ (1\ 1\ \ldots)\ \in \mathbf X $$


generates a dense (cyclic) subgroup--in other words, group $\ \mathbf X\ $ is monothetic.


