Let be $K \subset l_p$ closed set, with $(1\leq p<+\infty)$. Prove that $K$ is compact if and only if:
- $\forall n\geq1:\sup\{|x_n| : x=\{x_n\}_{n\geq1}\}<+\infty$
- $\lim_{m\rightarrow +\infty} \sup\{\sum_{n\geq m} |x_n|^p : x=\{x_n\}_{n\geq 1} \in K\}=0$
Here's what I was able to prove: I supposed $K$ compact and so, for the characterization of compact sets in a Banach space, $K$ must be bounded and closed. So, because $K$ is bounded is immediate that: $\sup\{|x_n| : x=\{x_n\}_{n\geq1}\}<+\infty$. I don't know how to prove the second point and I don't know to prove the viceversa, I guess I would have to prove that $K$ is bounded, and that comes easy from the first point, but how do I manage to create a collection of sets $M_\xi$ such that:
$$\forall \xi >0 \ \ \exists M_\xi \ subspace, \ dim(M_\xi)<+\infty \ : \ d(K, M_\xi) < \xi$$
A note: I define $lp$ as such: $$l_p=\{x=\{x_n\}_{n \geq 1} | \sum_{n \geq 1}|x_n|^p <+\infty\}$$