Let \(G\) be a first-countable topological group. Show that if \(G\) has a countable dense subset, or is Lindelöf, then \(G\) has a countable basis. [Hint: Let \(\ts{B_n}\) be a countable basis at \(e\). If \(D\) is a countable dense subset of \(G\), show the sets \(dB_n\), for \(d \in D\), form a basis for \(G\). If \(G\) is Lindelöf, choose for each \(n\) a countable set \(C_n\) such that the sets \(cB_n\), for \(c \in C_n\), cover \(G\). Show that as \(n\) ranges over \(\mathbb{Z}_+\), these sets form a basis for \(G\).]
Solution
Let \(B_1,B_2,\dots\) be a countable local basis at the identity \(e\).
Assume first that \(G\) has a countable dense subset \(D\). We claim the countable family \[ \{dB_n:d\in D,\ n\in\mathbb Z_+\} \] is a basis. Let \(x\in U\) with \(U\) open. Choose a neighborhood \(W\) of \(e\) such that \[ xWW^{-1}\subset U. \] Choose \(n\) with \(B_n\subset W\). Since \(D\) is dense, the open set \(xB_n^{-1}\) meets \(D\); choose \(d=xb^{-1}\) with \(b\in B_n\). Then \[ x=db\in dB_n, \] and \[ dB_n=xb^{-1}B_n\subset xB_n^{-1}B_n\subset xW^{-1}W\subset U \] after choosing \(W\) symmetric, or equivalently replacing it by a smaller symmetric neighborhood. Thus the displayed family is a countable basis.
Now assume \(G\) is Lindelöf. Shrink the local basis if necessary so each \(B_n\) is symmetric. For each \(n\), the translates \(\{gB_n:g\in G\}\) cover \(G\). Lindelöfness gives a countable set \(C_n\subset G\) such that \[ G=\bigcup_{c\in C_n}cB_n. \] The union over \(n\) of all \(cB_n\) is countable. To see it is a basis, let \(x\in U\) open. Choose \(n\) so small that \[ xB_nB_n\subset U. \] Pick \(c\in C_n\) with \(x\in cB_n\). Then \(c=xb^{-1}\) for some \(b\in B_n\), and therefore \[ cB_n\subset xB_nB_n\subset U. \] Hence \(G\) is second countable.