0

Let $c_0$ be the Banach space of all sequences converging to 0, equipped with the supremum norm.

How do the compact subsets of $c_0$ look like?

I could imagine that $K \subset c_0$ is compact if and only if K is closed, bounded and the sequences from K tend uniformly to 0, i.e., for all $\varepsilon > 0$ there is an $N \in \mathbb{N}$ such that $a_i < \varepsilon$ for all $i > N$ and all $a \in c_0$? If yes, how to prove this?

AlexE
  • 1,864

1 Answers1

2

Yes.

Since $c_0$ is complete, you only need to show total boundedness to show that a closed subset is compact. So let $K$ satisfy your criteria. Let $\epsilon>0$, and pick the corresponding $N$. Now $K_N=\{(a_1,\ldots,a_n)\colon a\in K\}$ is clearly a closed and bounded subset of $\mathbb{R}^N$. Cover it in a finite number of $\epsilon$-neighbourhoods. Take the centers of these, extend with zeros, take $\epsilon$-neighbourhoods of these in $c_0$, and you're done.

(I got interrupted while writing this, and now have to go to a two hour meeting. So this is woefully incomplete, but I am throwing it in there anyhow.)

Addition after thinking about it during the meeting: Assume $K$ is compact. Consider the functions $f_n\colon K\to\mathbb{R}$ given by $f_n(x)=\max_{k\ge n}|x_k|$. Then $f_n$ is continuous, $f_{n+1}\le f_n$, and $f_n\to0$ pointwise on $K$. By Dini's theorem, the convergence is uniform. So given $\epsilon>0$ there is some $N$ so that $f_n(x)<\epsilon$ for all $n\ge N$ and $x\in K$. But that means $x_k<\epsilon$ for all $k\ge N$ and all $x\in K$, which is your final condition.

Harald Hanche-Olsen
  • 31,960
  • @AlexE I see you accepted the answer already, but I felt the converse needed to be addressed as well, so I added it anyhow. (I already knew what to do, but the idea of using Dini's theorem came to me during the meeting.) – Harald Hanche-Olsen Dec 11 '13 at 14:33
  • Thanks, the argument via Dini's theorem is nice. But I think one can get this direction directly by contradiction: if the final condition is not satisfied, we take a sequence (of sequences) that witnesses this and then we can surely show that it has no convergent subsequence, contradicting the compactness. – AlexE Dec 11 '13 at 15:04
  • @AlexE You're certainly right. No telling which argument is shorter without writing them both out. Of course, a more elementary argument is often to be preferred, but I like shortcuts when I find a good one. – Harald Hanche-Olsen Dec 11 '13 at 15:55