Q) Suppose $(x_n)_{n\in \mathbb{N}}\in l^p(\mathbb{N}), 1\leq p<\infty$. Let
$$S=\{(y_n)_{n\in \mathbb{N}}: |y_n|\leq |x_n|, \forall n\in\mathbb{N}\}$$
Prove that $S$ is a compact subset of $l^p(\mathbb{N})$.
At a higher level, I think I need to show that for some sequence in $S$, the $l^p$ distance between the sequence and its limit converges to $0$ but not sure how to approach the question.
Let $(y_n^{(j)})$ be a sequence in $S$. Note that for each $n$ the sequence $(y_n^{(j)}:j\geq 1) $ is bounded and hence it has a convergent subsequence. By a diagonal procedure we can extract a subsequence $j_k$ such that $y_n=\lim_{k \to \infty} y_n^{(j_k)}$ exists for every $n$. We now claim that $(y_n^{(j)}) \to (y_n)$ in $l^{p}$. For this note that given $\epsilon >0$ there exist $N$ such that $\sum_{n=N} ^{\infty} |y_n^{(j_k)}|^{p} \leq \sum_{n=N} ^{\infty} |x_n|^{p} <\epsilon$. This also implies (by Fatou's Lemma, for example) that $\sum |y_n|^{p} <\infty$. Can you now show that $\sum |y_n^{j_k} -y_n|^{p} \to 0$?