Is every compact monothetic group metrizable?

Question

If $G$ is a compact (Hausdorff) topological group with a dense subgroup $H \cong \mathbb Z$, is it necessarily true that $G$ is first countable? This claim seems to be implicit in a paper that I am reading at the moment (the paper asserts the existence of a translation-invariant metric which induces the topology of $G$, which is equivalent by the Birkhoff-Kakutani theorem).

Answer 1

It seems the following.

Of course the answer is negative. Let $\Bbb T$ be the unit circle $\{z\in\Bbb C: |z|=1\}$ endowed with a topology from the complex plane $\Bbb C$ and with group multiplication of complex numbers. By Exercise 9.6.f from [AT], a compact abelian group $G$ is monothetic iff the dual group $G^*$ is isomorphic to a subgroup of the group $\Bbb T$ endowed with the discrete topology.

We can also directly show that a group $\Bbb T^\kappa$ is monothetic for each $\kappa\le\frak c$. Let $T=\{t_\alpha:\alpha<\kappa\}\not\ni 1$ be a set of irrational numbers such that a set $T\cup\{1\}$ is linearly independent over a field $\Bbb Q$ set. The family $T$ can be taken from a basis (containing $1$) of $\Bbb R$ considered as a linear space over the field $\Bbb Q$. Put $g=(g_\alpha)=(e^{2\pi t_\alpha i})\in \Bbb T^\kappa$. By [Pon, Example 65], for each sequence $t_{\alpha_1},\dots, t_{\alpha_r}$ of mutually different elements of $T$, each sequence $d_1,\dots, d_r$ of real numbers, and each positive number $\varepsilon$ there exist a sequence $n_1,\dots, n_r$ of integer numbers and an integer number $m$ such that $|mt_{\alpha_i}-d_i-n_i|<\varepsilon$ for each index $i=1,\dots, r$. This implies that the powers of $g$ are dense in the group $\Bbb T^\kappa$ endowed with the Tychonoff product topology.

References

[AT] Alexander V. Arhangel'skii, Mikhail G. Tkachenko, Topological groups and related structures, Atlantis Press, Paris; World Sci. Publ., NJ, 2008.

[Pon] Lev S. Pontrjagin, Continuous groups, 2nd ed., M., (1954) (in Russian).

Answer 2

This gives a counter-example if you don't ask for compactness.

Reference: Section 1 of http://matwbn.icm.edu.pl/ksiazki/fm/fm88/fm88117.pdf.

I will give the outline of what is done in the reference.

By a "metric" on an abelian group $G$, we will mean a function $m:G\to[0,\infty)$ satisfying $$(i)m(x)=0\iff x=0;\qquad (ii)m(x)=m(-x);\qquad (iii)m(x+y)\leq m(x)+m(y)$$ so a metric acts like a norm, and the equation $d(x,y)=m(x-y)$ gives the relation between invariant metrics (in the usual sense) and metrics in this new sense.

Let $T_A$ be the finest topology on $\mathbb{Z}$ such that $2^n\to 0$. For any sequence $p=(p_n)$ of strictly positive reals, the equation $$d_p(x)=\inf\left\{\sum_{n=0}^N|a_n|p_n:N\in\mathbb{N},a_n\in\mathbb{Z},x=\sum_{n=0}^Na_n2^n\right\}$ $$ defines a metric on $\mathbb{Z}$ such that $d_p(2^n)=p_n$. You can show that we can change the definition of $d_p$ and admit only $|a_n|\leq 1$ (use the fact that $p_n$ decreases).

Let $\mathcal{C}$ be the set of topologies induces by the metrics $d_p$ with $p$ non-increasing and for which $2^n\to 0$ in $(\mathbb{Z},d_p)$, or equivalently, $\lim_{n\to\infty}p_n=0$. Notice, in particular, that each member of $\mathcal{C}$ is Hausdorff and is a subtopology of $T_A$, so $T_A$ is Hausdorff.

Now, we can prove that $T_A$ is not metrizable: Let $d$ be any metric on $\mathbb{Z}$ such that $2^n\to 0$ in $(\mathbb{Z},d)$. If we show that $d$ cannot induce $T_A$, then we conclude that $T_A$ cannot be metrizable, because $2^n\to 0$ in $(\mathbb{Z},T_A)$. For every $x=\sum_{n=0}^Na_n2^n$, we have $$d(x)=d(\sum_{n=0}^N a_n2^n)\leq \sum_{n=0}^N d(a_n2^n)\leq\sum_{n=0}^N |a_n|d(2^n)$$ so, if we let $p_n=\max_{m\geq n}d(2^n)$ and $p=(p_n)_{n\in\mathbb{N}}$, we obtain $d(x)\leq d_p(x)$, so it suffices to show that $d_p$ does not induce $T_A$. For this, we will use some other metric of the form $d_q$.

Let $j\in\mathbb{N}$. Since $p_n\to 0$ (and $j$ is fixed), we can find $N(j)\in\mathbb{N}$ such that $\sum_{i=1}^j p_{N(j)+i}<1/j$. Also, we may assume that $N(j)+j<N(j+1)$. Notice that if $x(j)=\sum_{i=1}^j(-1)^i2^{N(j)+i}$, then this is the form of writing $x(j)$ as a sum of powers of $2$ with the least number of elements (and coeficient $0$, $1$ and $-1$). For $n$ between $N(j)$ and $N(j)+j$, we let $q_n=1/j$ and then extend $q=(q_n)$ to any non-increasing sequence. Then $d_p(x(j))<1/j$, so $x(j)\to 0$ in $(\mathbb{Z},d_p)$, but $d_q(x_j)=1$.

Finally, the topology $T_A$ is finer than the topology induced by $d_q$, and thus $d_p$ does not induce $T_A$.