Countability Axioms and Topological Groups (4.30.18)

James Munkres Topology, chapter 4.30, exercise 18:

Show in a first-countable topological group $G$ that second countability, separability, and the Lindelof condition are equivalent.

Proof: Let $\{B_n\}$ be a countable basis about $e$. We may presume $B_n⊇B_{n+1}$ for all $n$, and if necessary by setting $B_n'=B_n∩B_n^{-1}$ we may presume $B_n^{-1}=B_n$.

Suppose $G$ has a countable dense subset $D$, and that $U$ is a hood about $g$. Then since $g×e∈m^{-1}(U)$ and $\{gB_n\}$ is a basis about $g$, there exists $n$ such that $gB_n×B_n⊆m^{-1}(U)$. Choose $d∈D∩gB_n$ and we see $d=gb$ for some $b∈B_n$, hence $g=db^{-1}∈dB_n$, so $dB_n$ is a hood about $g$. Furthermore, given $c∈B_n$ we see $dc=gbc∈m(gB_n×B_n)⊆U$, so that $dB_n⊆U$. Therefore $\{dB_n~|~d∈D,n∈ℕ\}$ is a countable basis for $G$.

Suppose $G$ is Lindelof. Then for each $n∈ℕ$, there is a countable subset $D_n⊆G$ such that $∪_{d∈D_n}dB_n=G$. We show $D=∪D_n$ is dense in $G$; let $U⊆G$ be a hood about $g$. Then $gB_n⊆U$ for some $n$, and we also see $g∈∪_{d∈D_n}dB_n$ so write $g=db$ for some $d∈D_n⊆D$ and $b∈B_n$. But now $d=gb^{-1}∈gB_n⊆U$ so $d∈U$ and $D$ is a countable dense subset of $G$.$~\square$