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I need to show that $A = \{x\in \ell_2: |x_n| \leq \frac{1}{n}, n = 1,2,...\}$ is compact in $\ell_2$. I'm given the following hints:

  • First show that $A$ is closed.

  • Next, use the fact that $\sum_{n = 1}^{\infty}\frac{1}{n^2} < \infty$, to show that A is "within $\epsilon$" of the set $A \cap \{x\in \ell_2:|x_n| = 0, n\geq \mathbb{N}\}$.

Question: Why do these two steps/hints result in $A$ being compact? (I'm not so interested in how to perform these two steps, but more in why they would work).

Thanks in advance!

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    $\begingroup$ $l_2$ is certainly $not$ compact. $\endgroup$ – DanielWainfleet Nov 1 '17 at 8:21
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According to the Heine-Borel theorem a subset of a metric space is compact if and only if it is complete and totally bounded. The space $\ell^2$ is a complete metric space. Now according to this, a closed subspace of a complete metric space is also complete. Hence the usage of the first hint. The second hint is used for showing that $A$ is totally bounded.

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