# A countable tight topological group where every countable subset is metrizable

I am looking for an example of a topological group with countable tightness with the property then it is not metrizable, but every countable subset is metrizable but I cannot construct an example.

This question is inspired by my earlier question when-is-the-topology-generated-by-countable-subsets where someone gave me an example of a topological space with all of the above properties, namely the 1-point-compactification of an uncountable discrete set and now I was wondering if there are examples of topological spaces with these properties that carry a group strucuture.

Thanks in advance, Tom

## 1 Answer

Actually, Santi Spadaro almost gave the answer in a comment to your previous question: Let $X$ be the $\Sigma$-product of $2^{\omega_1}$ (where $2=\{0,1\}$), that is, the subset of $2^{\omega_1}$ of all points with at most countably many coordinates different from $0$. Of course $X$ is a subgroup of $2^{\omega_1}$.

Then, as explained in Henno Brandsma's answer here, $X$ is Frechet-Urysohn (and thus countably tight) and non-paracompact (and thus non-metrizable). But any countable subset is contained in some $2^{\alpha}$ for a countable $\alpha$, and the latter is metrizable since it is homeomorphic to the Cantor cube $2^\omega$.

• $2^\omega$ is the Cantor space, of course, not the Cantor cube. – Mathieu Baillif Feb 24 '15 at 9:54
• The Cantor space $2^\omega$ is a Cantor cube. – Ramiro de la Vega Feb 24 '15 at 14:18
• You can also let $X$ be the direct sum of $\omega_1$ copies of $2$. That is the set of all elements of $2^{\omega_1}$ with finite support. It is easier to show that this space is Frechet-Urysohn, and it is not first-countable (hence non-metrizable). – Ramiro de la Vega Feb 24 '15 at 14:21
• @Ramiro. I thought that term "Cantor cube" was reserved for $2^\kappa$ for uncountable $\kappa$. And you are right, the direct sum is a simpler example. – Mathieu Baillif Feb 24 '15 at 14:25
• Mathieu Baillif, Thank you for your answer and Ramiro de la Vega, thank you for the other example! :-) – Tom Mar 3 '15 at 9:51