# Prove that every infinite subset of the sequence space $l^2$ has a limit point in K

I'm having a hard time trying to hash this through - any help would be appreciated.

Let $$l^2$$ be the sequence space defined by $$\{\{a_n\}_{n=1}^{\infty}:a_n \in \mathbb{R}, \sum_n |a_n|^2 < \infty \}$$ Define $$K=\{x\in l^2 :|x_n| \leq \frac{1}{n} \ \forall n = 1,2,...\}$$. Show that every infinite subset of K has a limit point in K.

A suggested proof begins: let $$E \subset K$$ be and infinite subset and $$\{x_n\}_{n=1}^\infty \subset E$$ a sequence of distinct elements, where $$x_n = (x_{n,1},x_{n,2},...)$$ with $$|x_{n,k}|\leq \frac{1}{k}$$. I know I need to show convergence from here, and show that there exists a convergent sequence that is a limit point of E. However, not exactly sure where to go beyond defining that first sequence.

Thank you!

You don't need to show $$x_n$$ converges, you need to show a subsequence of $$x_n$$ converges. The entire question could be rephrased as "show $$K$$ is a compact subset of $$\ell_2$$".
First note that $$K$$ is a bounded set, if $$x \in K$$ then $$\lVert x \rVert_2 \leq \sum_k \frac{1}{k^2} < \infty$$. Then by applying a diagonalisation trick, given a sequence $$(x_n) \subset K$$, we can find a subsequence $$(x_{n_m})$$ which converges in every component, meaning $$x_i = \lim_{m \to \infty} x_{n_m, i}$$ exists for all $$i$$. This is a common trick when working with infinite sequences, if you're unfamiliar with this argument please say.
It should be clear that $$x_i \leq \frac{1}{i}$$ hence $$x = (x_1, x_2, \ldots) \in K$$. So we need to prove $$\lVert x_{n_m} - x \lVert_2^2 \to 0$$ to prove convergence in $$\ell_2$$. To do this we exploit the definition, $$\sum_n \frac{1}{n^2} < \infty$$ so for all $$\epsilon > 0$$ we can pick $$I$$ such that $$\sum_{i \geq I} \left(\frac{2}{n}\right)^2 < \frac{1}{2} \epsilon$$. Then \begin{align*} \lVert x_{n_m} - x \rVert_2^2 &= \sum_{i=1}^{\infty} \lvert x_{n_m, i} - x_i \rvert^2\\ &\leq \sum_{i=1}^I \lvert x_{n_m, i} - x_i \rvert^2 + \sum_{i=I}^{\infty} \left(\frac{2}{n}\right)^2 \\ &< \sum_{i=1}^I \lvert x_{n_m, i} - x_i \rvert^2 + \frac{1}{2}\epsilon \\ &\to \frac{1}{2} \epsilon \quad \text{as m \to \infty} \end{align*} We could safely interchange the limit $$m \to \infty$$ and the sum since the sum is now a finite one. This result holds for all $$\epsilon > 0$$ hence $$\lim_{m \to \infty} \lVert x_{n_m} - x \rVert_2^2 \to 0$$ as required.