Compact set in normed vector space of infinite dimension

Question

I know that in metric spaces, compact set are set such that every sequence has a convergent subsequence. As a surprising fact of a question I asked here, in infinite dimension, balls are not compact.

• 1) So how looks compact sets in infinite dimension normed space ? I guess that they must be at least closed (since convergent sequence must converge in the set). But in other way, unit ball is closed (and bounded), thus maybe compact set are not closed. So how do they look ?

• 2) In weak topology, I know that since there are less open set, there are more compact set. But if there are less open set, necessarily there are less closed set, right ? And thus, it has less compact set, no ? If compact set are at least not closed, what could be there interest ?

Answer 1

(1) Compact sets are still closed. However, they can't look like balls; they have to be very "thin", almost finite dimensional.

If you think about a Hilbert space like $\ell^2$, you might observe that the following sets are compact:

• the set of all $x$ where $|x_i| \le 1$ for $1 \le i \le 100$, and $x_i = 0$ for $i > 100$. This is really finite dimensional because all the action takes place in the first 10 coordinates.

• the set of all $x$ where $|x_i| \le 1/i$. This is "almost" finite dimensional because, in some sense, for any $\epsilon$, all but $\epsilon$ of the action takes place in a finite number of coordinates.

(2) There's no contradiction between having more compact sets and fewer closed sets. It just means that, very roughly, a larger "fraction" of the closed sets will be compact. When you go from the norm to the weak topology, you'll lose some closed sets, but you won't lose any of the ones that were compact in the norm topology; those sets will still be closed (and compact) in the weak topology. (Another useful fact is that you won't lose any of the convex closed sets either).

Answer 2

1) In metric spaces, a set $E$ is closed $\iff$ all convergent sequences of $E$ converge in $E$. As you see, to be closed you don't need to have a convergent subsequence. The only thing is if it converge, it must converge in the set (take for example $x_n=n$ in $[0,\infty [$, the set $[0,\infty [$ is closed but $(x_n)$ has no convergent subsequence).

So as you said, a necessary condition to be compact is to be closed, but unfortunately it's not sufficient. Even being closed and bounded is not enough. But as you can prove, it's necessary. i.e. $\{Compact\}\subsetneq \{bounded\ closed\}$. Indeed, it's easy to prove that unbounded set won't be compact.

2) Notice that compact set must be closed is in metric space only. The weak topology is unfortunately not metrizable, so the property of being closed doesn't hold any more. The key here is : to check that $K$ is compact, you'll have less covering of $K$ by open set, and thus more chance to have all covering that will admit a finite sub-covering.

Just for illustration. Let $(X,\mathcal T)$ a topological set. If $\mathcal T=\{\emptyset, X\}$ then all set will be compact and if $\mathcal T$ is the discrete topology, only finite set will be compact.