# If G is a separable metrizable group with at least two elements, then $G^{\omega_1}$ is a separable Fréchet group which is not metrizable.

(Assuming $\mathfrak{p}>\omega_1$) If G is a separable metrizable group with at least two elements, then $G^{\omega_1}$ is a separable Fréchet group which is not metrizable.

This is stated without proof in the survey article by Justin Tatch Moore and Stevo Todorcevic in Open Problems in Topology 2 titled The metrization problem for Fréchet groups.

I've never studied topological groups before and I have some questions.

How am I supposed to interpret $G^{\omega_1}$? Are its elements finite support $\omega_1$-sequences (almost always $1_G$) or just all $\omega_1$-sequences?

It would also be very useful to me if you could help me verify the claim. It has been suggested to me that this claim might actually be false and a mistake by the authors. (But this might be because of a wrong interpretation of $G^{\omega_1}$.) I'm wondering which of the three properties (separable, fréchet, non-metrizable) are always true and which need the assumption of $\mathfrak{p}>\omega_1$.

Thank you.

EDIT: I've found the following claim which is probably what Moore and Todorcevic intended to convey with their example:

Assuming $\mathfrak{p}>\omega_1$, every countable dense subgroup of $2^{\omega_1}$ is Fréchet and not metrizable.

This subgroup would be Fréchet and separable (both simply because it's countable), but how can I verify it's not metrizable?

• I would assume it means all $\omega_1$-sequences, with the product topology. All the properties except it being Fréchet are then standard facts (a product of $\leq\mathfrak{c}$ separable spaces is separable, and a nontrivial uncountable product cannot be metrizable). But unless I'm making a mistake, it seems not to be Fréchet: if $a,b\in G$ are two distinct elements of $G$, then the set of functions $\omega_1\to G$ that are $a$ at countably many points and $b$ everywhere else has the constant function $a$ as a limit point, but not the limit of any sequence. – Eric Wofsey Nov 4 '16 at 2:48
• @EricWofsey I edited with some new information! Thank you for your comment, I agree your proof of it not being Fréchet seems to be correct. – JKEG Nov 4 '16 at 12:55