Alex Fekken (5/19/2013)
You made me doubt myself for a moment, but:
You are right. I should be more careful when speculating without references. AC implies that there are bounded infinite point sequences that have no limit points in any infinite-dimensional normed space, so it can't be compact. I wish I'd been aware of that 46 years ago!
http://planetmath.org/compactnessofclosedunitballinnormedspaces
[cool math]Pretty sure the same is true of any metric space: cover the ball with open neighbourhoods of the endpoints of a basis of unit vectors where each such neighbourhood covers only one such endpoint, as well as any open sets that you need to cover the rest of the ball but that do not cover those endpoints. If you have an infinite number of dimensions then you won't be able to refine the cover to a finite one because each endpoint is covered by only one open set.[/cool math]
That probably works for infinite dimensional metric vector spaces even if the metric isn't induced by a homogenous norm, so if you have the definition of dimensionality which requires a vector space it's fine. I don't recall whether that definition of dimensionality was generally accepted though, there used to be many other definitions, several of which didn't even require a metric, let alone a coordinate system. I'm not sure whether it works for non-vector space metric spaces (if we use a definition that allows them to have infinite dimensionality, and maybe a field other than the reals for the distance field). That's probably too difficult question for my (extremely rusty) knowledge of topology.
Tom