Compactness in $\ell ^1$. Is $K := \{ (z_n)_n \in \ell^1 : |z_n| \leq |x_n| \}$ compact? I'm having some trouble with the following:
Consider the following set $K := \{ (z_n)_n \in \ell^1 : |z_n| \leq |x_n| \}$ for a given $(x_n)_n \in \ell^1$. Is $K$ a secuencially compact space?
This is my thoughts so far:
I suspect $K$ is in fact compact. We want to see that given a $(\xi_k)_k \subset \ell^1$ we can find a convergent subsequence. So let $\xi_k= (z^k_n)_n$ and $(\xi_k)_k \subset K$. For a fixed $n$ we have that $(z^k_n)_k \subset \bar B_{|x_n|}(0) \subset \mathbb{R} $, which is compact. It follows then that $(z^k_n)_k$ has a partial convergent subsequence. Let then, $(z^{k'}_n)_{k' \in I(n)}$ where $I(n) \subset \mathbb{N}$ is an index set, denote such convergent subsuccession. My idea is that if we can see that $|\cap_n I(n)| = \infty$ then $(\xi_{k'})_{k'}$ with $k' \in \cap_n I(n)$ would be a convergent subsequence of $(\xi_k)_k$. I'm a bit stuck on how to proceed from here, suposing that this leads somewhere interesting.

As you have pointed out, we have for fixed $n$ that the sequence of the nth entry of our sequence is bounded and thus admits a converging subsequence. Using a Cantor diagonal argument we can produce a subsequence $(\xi^{k_l})_l$ such that for every fixed $n$ the sequence of the $n$th entries converges. Set $y= (y_n)_n=(\lim_{l\rightarrow \infty} \xi^{k_l}_n)_n$.
Note that
$$ \Vert y -\xi^{k_l} \Vert
\leq \sum_{j=1}^N \vert y_j - \xi^{k_l}_j \vert + \sum_{j\geq N+1} 2\vert x_j\vert . $$
From this we see, that our subsequence converges in $\mathcal{l}^1$.
