Compact set in a Hilbert Space

I'm doing some exercises, but there is something in one of the questions I don't quite understand, so I'm really hoping someone might be able to clarify. The question is as follows: Let $\mathcal H$ be a Hilbert Space with inner product $\lt\cdot,\cdot\gt$, and corresponding norm $\|\cdot\|$. Furthermore there is the set $B=\{x\in\mathcal H |\|x\|\le1\}$.
a) Show, that B is closed and bounded$^1$ regarding (the metric corresponding to) $\|x\|$.
b) Show, that if $\mathcal H$ doesn't have a finite dimension, then B isn't compact regarding (the metric corresponding to) $\|x\|$. (Hint: Look at a countable orthonormal system $\{e_n|n\in\mathbb N\}$ in $\mathcal H$.
c) Give an example of a Hilbert Space $\mathcal H$, where B isn't compact.

What I don't understand is, if I have to show in a) that B is compact, how can it suddenly be not-compact in b)? Where do we assume that $\mathcal H$ has a finite dimension in a)? And finally, in c) - haven't we just, in b) given an example - the example being all Hilbert Spaces which does not have a finite dimension?

$^1$ This is the same as being compact, right?

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Closed and bounded does not imply compact - but compact implies closed and bounded. (I've made the same mistake many times.)

a) Of course if you could prove that $B$ is compact then you'd achieve this...but this claim is false!

b) So up to here is consistent. You don't assume $\mathcal{H}$ has finite dimension in $A$, since again closed and bounded is not the same as compact.

c) Sure, that's a valid example. Maybe you want to try considering the question for finite dimensions?

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a) If B was always compact then you'd run into problems with compact operators, because the identity is never a compact operator. –  neuguy Dec 10 '12 at 4:22

a) No, this is not in general true. For example, if you give $\mathbb{Z}$ the discrete metric, the whole space is closed and bounded but not compact. I think you can actually solve part a) by using the continuity of $\|\cdot\|$ and the fact that you're ball is the preimage of $[0,1]$.

b) You should try as hard as you can. If you need to, you can peek here Is any closed ball non-compact in infinite dimensional space?.

c) Yes, you are correct. Take your pick, $L^2$ for example.

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@JonasMeyer You are of course correct--my head must have been elsewhere (I was thinking of Banach spaces). –  Alex Youcis Dec 10 '12 at 4:18