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Is every first countable profinite group actually second countable?

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up vote 6 down vote accepted

Although this question has been answered already by Masked Avenger, it seems like a good idea for me to explain why every first-countable Hausdorff group is metrizable. If $G$ is a Hausdorff topological group and $U$ is an open neighborhood of the identity $e$, then let $R_{U}$ be the binary relation on $G$ where $(x,y)\in R_{U}$ iff $xy^{-1}\in U$. Then $G$ becomes a Hausdorff uniform space with a uniformity generated by the relations $R_{U}$ where $U$ ranges over all open neighborhoods of the identity $e$. Furthermore, it is well known that a uniform space $(X,\mathcal{U})$ is induced my a metric if and only if $\mathcal{U}$ is generated by countably many relations (a proof of this fact is given in this answer). Therefore, if a $G$ is a first countable Hausdorff topological group, then the uniformity generated by the relations of the form $R_{U}$ is metrizable. Therefore, we conclude that every Hausdorff first countable topological group is metrizable.

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J. Wilson, Profinite groups, 4.1.3: a profinite group is metrizable iff it is a countable inverse limit of finite groups. Any first-countable, Hausdorff group is metrizable. So, yes, every first-countable profinite group is second-countable, this is why they are usually called "separable".

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