Let \(G\) be a locally compact, connected topological group. Show that \(G\) is paracompact. [Hint: Let \(U_1\) be a neighborhood of \(e\) having compact closure. In general, define \(U_{n+1} = \overline{U}_n \cdot U_1\). Show the union of the sets \(\overline{U}_n\) is both open and closed in \(G\).]
This result holds without assuming \(G\) is connected, but the proof requires more effort.
Solution
Because \(G\) is locally compact Hausdorff, choose a symmetric open neighborhood \(U\) of the identity \(e\) whose closure \[ K=\overline U \] is compact. Replacing \(U\) by \(U\cap U^{-1}\) if necessary preserves compact closure and makes \(K=K^{-1}\).
Set \[ H=\bigcup_{n\ge1}K^n. \] Each \(K^n\) is compact, since multiplication \(G^n\to G\) is continuous. Also \(H\) is a subgroup: \(e\in K\), products send \(K^mK^n\) into \(K^{m+n}\), and symmetry of \(K\) gives closure under inverses. Since \(U\subset K\subset H\), the subgroup \(H\) contains an open neighborhood of the identity, hence \(H\) is open. Every open subgroup of a topological group is also closed, since its other cosets are open. Connectedness of \(G\) therefore forces \(H=G\).
Thus \[ G=\bigcup_{n\ge1}K^n \] is \(\sigma\)-compact. Hausdorff topological groups are regular, and every regular \(\sigma\)-compact space is paracompact by E-KVFCT. Hence \(G\) is paracompact.