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

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?

You have to replace $$x_n \leq \frac 1 n$$ by $$|x_n| \leq \frac 1 n$$. Otherwise the set is not even bounded.
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 $$k_1 such that $$\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.