Proposition: Let be a Hausdorff topological group. Then a subgroup of that is locally compact (in the subspace topology) is moreover closed. In particular, every discrete subgroup of is closed.
I've been struggling over the proof of this all day. I have been reading a proof in Fourier analysis by Ramakrishnan & Valenza and I understand it all except for this one point. It starts off by using local compactness of and letting be a compact neighbourhood of the identity element in the subspace topology of which is fine. Then it concludes therefore there is a closed neighbourhood of in such that . How on earth does one achieve this? I mean, its easy to find a closed set with that, but how do I also make sure is a neighbourhood of ?