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Show $A = \{x \in l_2: x_n \leq \frac{1}{n}$, $n = 1,2,\ldots\}$ is totally bounded in $\ell_2$ using sequences

I know that $A$ will be totally bounded if every sequence in $A$ has a Cauchy subsequence.

Can someone please tell me how can I get a sequence in $A$ that is not Cauchy?

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You have to replace $x_n \leq \frac 1 n$ by $|x_n| \leq \frac 1 n$. Otherwise the set is not even bounded.

Use a diagonal procedure to show that every sequence has a convergent subsequence:

Let $x^{1},x^{2},...$ be a sequence in $A$. Since the first coordinates are bounded we can extract a convergent subsequence. Now look at the second coordinates along the subsequence, and so on. By a diagonal procedure we get $k_1<k_2<...$ such that $\lim_{l \to \infty} x^{k_l}_j=x_j$ exists for each $j$. Now $|x_l| \leq \frac 1 l $ and $(x_l) \in \ell_2$. Also $x^{k_l} \to x$ in the norm of $\ell _2$. [Can you show this?] We have proved that any sequence in $A$ has subsequence which converges in $\ell_2$. Hence $A$ is totally bounded.

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