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This question is motivated by this MO-problem 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?

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.

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.

However the subgroups $\mathfrak E$ and $G$ of $\ell_2$ are not monothetic.

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    $\begingroup$ Is there a typo in the title? $\endgroup$ – Wlod AA Jun 17 '18 at 5:09
  • $\begingroup$ @WlodAA Yes, there was a typo (now correcetd by Alon Amit, I hope). Tahnk you for noticing that. $\endgroup$ – Taras Banakh Jun 17 '18 at 13:04
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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).

$\mathbf{\ell^2}$-product of arbitrary metric spaces

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.


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.

An example of a monothetic group

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.

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