I'm trying to understand Rudin's proof of Pontryagin duality theorem, but I still haven't undersood an argument. (Fourier analysis on groups, p29)
Let $G$ be a group and denote $\Gamma =\widehat{G}$ his dual group. Every $x\in G$ can be considerated as an element $\alpha (x)$ of $\widehat {\Gamma}$ where $\alpha (x)(\gamma )= \gamma (x)$ for all $\gamma \in \Gamma $. It as already be shown that $\alpha $ is a homeomorphism onto his image, so the topology of $\alpha (G)$ is locally compact. But this topology is the topology inhérited as a a subspace of $\widehat{\Gamma }$. We have to show that $\alpha (G)$ is closed. Let $\gamma _0$ be a point of the closure of $\alpha (G)$, and $U$ an open neighbourhood of $\gamma _0$ such that $\overline{U}$ is compact. Why must $\alpha (G)\cap \overline{U}$ be compact ?
In general, a locally compact subset of a compact set is not closed.
It's probably a silly question, but any help would be appreciated.
It seems the following. It is well known and easy to show the next
Lemma. [Eng, Theorem 3.3.9] Every locally compact subspace $M$ of a Hausdorff space $X$ is an open subset in the closure $\overline{M}$ of the set $M$ in the space $X$.
Now, since the space $\alpha (G)$ is locally compact, it is an open subgroup of the group $\overline{\alpha (G)}$. Since each open subgroup of a topological group is closed (because its complement is a union of cosets, which are open as homeomorphic copies of the subgroup), $\alpha (G)$ is closed in $\overline{\alpha (G)}$. But $\alpha (G)$ is dense in $\overline{\alpha (G)}$, so $\alpha (G)=\overline{\alpha (G)}$.
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.