Seems right to me, and I don't see anything that's redundant.
Hey everyone, I think I am making the following WAY too difficult. Does anyone have insight into A) whether the following is even correct or B) an easier way to do it (as I am sure there is one).
Problem: Let be a locally Euclidean space of dimension . Call an open subset of a Euclidean ball if for some open . Prove that has an open base of Euclidean balls.
Proof: Let be arbitrary and let be the guaranteed Euclidean ball containing with the necessary homeomorphism. Let (each of these balls are open in for obvious reasons) and . Clearly contains open subset of , but since is in fact open we see that each element of is open in . So, set . By previous comment this is a class of open subsets of . Also, is a homeomorphism and so is in fact a collection of Euclidean balls. It remains to show that it's a base.
So, let be arbitrary and any neighborhood of . Thus, keeping the notation of the previous paragraph we have that is an open subspace of and so is an open subset of . So, there exists some . It follows that and . The conclusion follows. .
Comments?