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?
Best Answer
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?