Compactness in Hilbert space $\ell^2$

Question

If we have a compact set $A$ contained in $\ell^2$ (Hilbert Space), how can we show that $A$ is closed and bounded?

In $l^2$ closed and bounded is not equivalent with compact. – Oct 15 '18 at 20:40 — , Oct 15 '18 at 20:40
The implication does hold @Math_QED. – Henno Brandsma Oct 15 '18 at 21:12 — Henno Brandsma, Oct 15 '18 at 21:12

score 2 · Answer 1 · answered Oct 15 '18 at 21:11

This holds in any metric space, closedness because of Hausdorffness and boundedness by considering covers like $\{B(x,n): m \in \mathbb{N}\}$ for a fixed $x$.

Closed and bounded does not imply compactness, the unit ball is a counterexample (what limit can a subsequence of $(e_n)$ have?).

Compactness in Hilbert space $\ell^2$

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