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!


1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .