Let $ G $ be a finite dimensional real Lie group, and take a bounded ball $ B_R(e) \subset G $ in it, coming from the Riemannian metric, which itself is induced from an inner product on $ \mathfrak{g} = \text{Lie}(G) $.
I want to show, that such balls are totally bounded. For small $ R $, this is trivial, because $ G $ is locally compact. For general $ R $ it seemed to be true to me, but I did not find an answer to it. I had trouble to work with the dimension of $ G $. What I mean with this: In $ \mathbb{R}^n $ we need (about) $ (R/ \epsilon)^{\dim(G)} $-many $ (-\epsilon, \epsilon)^{\dim(G)} $-boxes to cover $ B_R(e) $. I wasn't able to carry this argument over to the Lie group $ G $.
Am I overseeing something easy? Or is it not even true?
Ok, I got the expected positive answer, but with an overkill argument. The generalized Hopf-Rinow theorem applies here, which states, that if a length space is complete and locally compact (which all three are satisfied by a Lie group $ G $), then the space is in particular proper, i.e. closed and bounded sets are compact. See for example:
https://www.math.ethz.ch/~lang/LengthSpaces.pdf
But there must be a shorter and nicer argument; anybody an idea?