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I'm trying to understand the proof of all compact topological groups are $\mathbb{R}-$factorizable, proof by Tkatchenko. I'm assuming all groups to be Tychonoff.

I defined the quotient group $\frac{G}{N}$ where $N$ is the coset containing the identity. I've already proved that $N$ is closed, hence $G/N$ is compact, Hausdorff. And I also proved that $\frac{G}{N}$ is metrizable. My only problem is to understand why $\frac{G}{N}$ is second countable. I need this to conclude that the group $G$ is $\mathbb{R}-$factorizable, maybe there's some equivalence I'm not seeing.

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    https://math.stackexchange.com/questions/573787/compact-metric-spaces-is-second-countable-and-axiom-of-countable-choice – Moishe Kohan Apr 05 '24 at 16:33

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Every compact metrizable space is second countable. See this post: Does every compact metric space have a finite basis?.

To explain the argument intuitively, in a metrizable space, we are allowed to say that the topology is induced by a norm. Thus, every open set of the topology is a ball centered around a point of $\frac{G}{N}$. Take a cover of the space made by balls of radius $r$, with $r < 1$. Take a finite subcover, and then do the same for balls of radius $r^2$. Repeat this process, taking unions as you go. Every open ball of the topology can then be expressed as the unions of these smaller balls (the base constructed by this process consists of arbitrarily small open balls).