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I am going through Carothers Real Analysis to study for quals, and came across this question.

Show that the set $A = \{x \in \ell_2 : \lvert x_n \rvert \leq \frac{1}{n}, n \in \mathbb{N}\}$ is compact in $\ell_2$. Hint: 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 $A \cap \{x \in \ell_2 : \lvert x_n \rvert = 0, n \geq N\}$ for some $N \in \mathbb{N}$.

Now I can prove what the hint says to prove. What I do not understand is how on earth this shows that $A$ is compact. This question comes at the beginning of the compactness chapter and all I have at my disposal is that a set is compact iff it is totally bounded and complete iff every sequence has a convergent subsequence, and that if $A \subset M$ and $A$ is compact, then $A$ is closed in $M$, and that if $M$ is compact and $A$ is closed, then $A$ is compact.

Aphyd
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  • Closed + totally bounded = compact. And you've shown both! – Matematleta Sep 10 '19 at 02:34
  • How is A intersect that set totally bounded? That set is basically Rn. Ah! But it is intersected with A, so its entries are less than 1/n, so it is totally bounded! – Aphyd Sep 10 '19 at 02:41
  • I'm actually wondering the same thing. That shows that $A$ is covered by the $\epsilon$-neighborhood of some compact set $C_\epsilon.$ Is that sufficient to show $A$ is totally bounded? – Brian Moehring Sep 10 '19 at 02:51
  • I was thinking it would be possible that A intersect that set is totally bounded, and then adding the tails would still make A totally bounded. But I haven't actually done the proof yet – Aphyd Sep 10 '19 at 02:53
  • this question is marked as a duplicate so I can't provide an answer. But here is the idea: take ${x \in \ell_2 : \lvert x_n \rvert = 0, n \geq N}$ and extract an $\epsilon$-net: for each component $n<N$, cover the intervals $[-1/n,1/n]$ by a finite number of $\epsilon$ balls, and locate the center of each one. For each component, you get a finite number of points, and then you collect them into a finite number of sequences in ${x \in \ell_2 : \lvert x_n \rvert = 0, n \geq N}$. By construction, each element of $A$ is at an $\epsilon$ distance from one of those points. – Matematleta Sep 10 '19 at 03:14

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