An infinite product of locally metrizable spaces is locally metrizable(in the product topology)?

Question

Locally metrizable means that each point has a neighborhood that is metrizable in the subspace topology.

I think it can be proved that a finite product of locally metrizable spaces is locally metrizable. But how about the infinite case(both countable and uncountable)? If it isn't true, are there any counterexamples?

Answer 1

Let $X$ be any locally metrizable space that is not globally metrizable (e.g., $\omega_1$). Then any infinite power $X^I$ is not locally metrizable. Indeed, any nonempty open set must contain a homeomorphic copy of $X$ (since some coordinate must be unconstrained in any basic open set), and thus cannot be metrizable.

Answer 2

If $X^I$ is locally metrisable, and $f \in X^I$, then $f$ has a metrisable neighbourhood of the form $U=\prod_{I \in I} U_i$ where all $U_i = X$ for $i \notin F$ and all other $U_i$ are open and $F \subseteq I$ is finite. So $X$ is metrisable if $I$ is infinite, as $X$ is always an embeded subspace of $U$ in that case. Also $U$ is metrisable implies $I$ is countable (as an uncountable product of non-trivial (more than one point) spaces is never metrisable). So $I$ is either countably infinite and $X$ is metrisable, or $I$ is finite and $X$ is locally metrisable. These are necessary and sufficient.