# Prove $l_1$ space is complete using: 'every sequence has a converging subsequence'

As in this post, I am trying to prove that: $$l_1 := \{x \in (x_n)_{n=1}^{\infty} | x_n \in \mathbb{R}, \sum_{n=1}^{\infty} |x_n| < \infty \}$$

equipped with $$||\cdot||_1$$, is complete.

Since I've been able to do this using Cauchy (as in the link above), I'm now trying, for practice, to do this using the compact-equivalent property that every sequence has a converging subsequence, but I'm stuck with completing it since I'm not sure how to show that I end up with a subsequence of the original one that converges (using epsilon preferably and not limit).

Here's what I have so far:

Let $$(\bar{x}^k)_{k\in \mathbb{N}} \subset l_1$$. Since $$(\mathbb{R}, |\cdot|)$$ is a complete space, then $$(x^k_1)_{k \in \mathbb{N}}$$ has a converging subsequence $$(x^{k_{l(1)}}_1)_{l(1)\in \mathbb{N}}$$. Denote its limit by $$x_1^0$$.

The sequence $$(\bar{x}^{k_{l(1)}})_{l(1)\in \mathbb{N}}$$ taken at index 2, i.e., $$(x^{k_{l(1)}}_2)_{l(1)\in \mathbb{N}}$$ also has a converging subsequence. Denote its limit by $$x_2^0$$.

Generally, we obtain that the sequence $$(\bar{x}^{k_{l(n-1)}})_{l(n-1)\in \mathbb{N}}$$ taken at index n, i.e., $$(x^{k_{l(n-1)}}_n)_{l(n-1)\in \mathbb{N}}$$ also has a converging subseqence. Denote its limit by $$x_n^0$$.

Now what? I can't really take $$n$$ to infinity since then the following claim breaks with the $$sup$$ not being defined:

Let $$\epsilon > 0$$. Then for every $$N> \sup\{\{N_n\}_{n\in \mathbb{N}}\}$$ where $$N_n$$ is such that for every $$n > N_n$$ we have $$|(x^{k_{l(n)}}_n)_{l(n)\in \mathbb{N}}-x_n^0| < \frac{\epsilon}{2}\frac{1}{2^n}$$ and the claim follows.

Any help would be much appreciated!

You’d be on the right track, what you’d take there is the diagonal sequence. But this won’t lead you anywhere, as your space is not compact. It does not work for $$\mathbb R$$ either: Take the sequence $$x_n = n$$. Clearly this has no convergent subsequence.