Let \(H\) be a subspace of the topological group \(G\). Show that if \(H\) is also a subgroup of \(G\), then both \(H\) and \(\overline{H}\) are topological groups.
Solution
Let \(m\colon G\times G\to G\) and \(i\colon G\to G\) be multiplication and inversion, which are continuous.
1. If \(K\subseteq G\) is a subgroup, then \(K\) with the subspace topology is a topological group.
By E-R3NOE, the product topology on \(K\times K\) is the subspace topology from \(G\times G\). The maps \(m|_{K\times K}\) and \(i|_K\) are restrictions of continuous maps to subspaces, and they take values in the subspace \(K\) because \(K\) is a subgroup; a continuous map with image in a subspace is continuous into that subspace.
2. \(\overline H\) is closed under inversion.
\(i\) is continuous and \(i\circ i=\operatorname{id}_G\), so \(i\) is a homeomorphism and \(i(\overline H)=\overline{i(H)}=\overline H\).
3. \(\overline H\) is closed under multiplication.
Let \(x,y\in\overline H\) and let \(W\) be an open neighborhood of \(xy\). By continuity of \(m\) at \((x,y)\) there are open \(U\ni x\) and \(V\ni y\) with \(UV\subseteq W\). Choose \(h_1\in U\cap H\) and \(h_2\in V\cap H\); then \(h_1h_2\in W\cap H\). Hence every neighborhood of \(xy\) meets \(H\), and \(xy\in\overline H\).
4. Q.E.D.
Step 1 applies to \(H\). The identity lies in \(H\subseteq\overline H\), so by steps 2 and 3 \(\overline H\) is a subgroup, and step 1 applies to \(\overline H\).