Describe all the compact subsets of $\ell^{1}$ -- General Result Proved, Example Needed. The space $\ell^{1}$ the space of all infinite sequence $\mathbf{x}:=(x_{1},x_{2},x_{3},\cdots)$ such that the infinite sum of the coordinate is absolutely convergent. That is, $\sum_{i=1}^{\infty}|x_{i}|<\infty$.
We give this space a metric defined as $$d(\mathbf{x},\mathbf{y}):=\sum_{i=1}^{\infty}|x_{i}-y_{i}|,$$ and I want to study on the compactness of this space and its subsets.
I read several online notes and post in the stackexchange, but what I got was mostly the subsets that are not compact. For example, http://math.stanford.edu/~ksound/Math171S10/Hw7Sol_171.pdf this note shows that $\ell^{1}$ itself is not compact. This post Closed and bounded but not compact subset of $\ell^1$ shows that even closed and bounded subset of $\ell^{1}$ is not compact (so our adorable closed unit ball is not compact).
The only thing I got is the page 24 of this note: https://www.math.kit.edu/iana3/~schnaubelt/media/fa14-skript.pdf, but it only gives a sufficient and necessary condition for a subset $K\subset\ell^{p}$ to be relatively compact in $\ell^{p}$, not compact.
Is there a way to describe the compact subsets of $\ell^{1}$? Or is there any sufficient (and/or necessary) condition of a subset of $\ell^{1}$ to be compact?
Thank you!
Edit 1:
As commented by Alessandro, I have known the sufficient conditions of relative compact. A subset $K$ that is relatively compact in $\ell^{1}$ has the closure $\overline{K}$ compact in $\ell^{1}$. Therefore, if I require additionally the set to be closed, then the closure is the set itself and thus the set is compact in $\ell^{1}$.
Therefore, combining the Proposition 1.45 in the page 24 of the note I linked above. We have the following proposition:

Proposition. Let $p\in[1,\infty)$. A set $K\subset\ell^{p}$ is compact if and only if it is closed and bounded, and $$\lim_{N\rightarrow\infty}\sup_{(x_{j})\in K}\sum_{j=N+1}^{\infty}|x_{j}|^{p}=0.$$

However, I don't know if such a set truly exists. Is it possible to construct a set $K\subset\ell^{p}$ such that it satisfies all these requirements?
Edit 2:
As mentioned in the above edition, we have found a general sufficient condition. However, I am not sure if such a set really exists.
As commented by "Kavi", one such set can be $\{\mathbf{0}\}$. Indeed, it is clearly bounded. Any singleton is closed with respect to any metric space, proved here: are singletons always closed?. This set contains only zero sequence, so clearly it satisfies $$\lim_{N\rightarrow\infty}\sup_{(x_{j})\in K}\sum_{j=N+1}^{\infty}|x_{j}|^{p}=0.$$
Therefore, $\{\mathbf{0}\}$ is a compact subset of $\ell^{p}$.
However, is this the only set? Are there any other examples? "Kavi" commented that $\{\mathbf{0}\}$ is the only linear subspace that is compact in $\ell^{p}$, why is this true? Does this mean $\{\mathbf{0}\}$ is the only compact subset? why?
Thank you!
 A: 
Is there a way to describe the compact subsets of $\ell^{1}$? Or is there any sufficient (and/or necessary) condition of a subset of $\ell^{1}$ to be compact?

You already answered that question with the proposition you gave.
I do not think that it can get much nicer than that.
Note that a closed and bounded are important properties of compact sets
in general.

Therefore, $\{\mathbf{0}\}$ is a compact subset of $\ell^{p}$.
However, is this the only set? Are there any other examples?

Yes, there are many other examples.
For example,
any finite dimensional subspace intersected with the closed unit ball of $\ell^1$
is compact (this follows from the description of compact sets
in finite dimensional spaces).
One can also construct compact sets that are not subsets of finite dimensional subspaces,
for example
$$
\{ x\in\ell^1 | x_i\in [0,1/i] \;\forall i\in\Bbb N\}.
$$
This can be verified using the proposition that you mentioned.

"Kavi" commented that $\{\mathbf{0}\}$ is the only linear subspace that is compact in $\ell^{p}$, why is this true?

If you have any other linear subspace, then this subspace will not be bounded,
and therefore cannot be compact.

Does this mean $\{\mathbf{0}\}$ is the only compact subset? why?

No, this is not the only compact subset, see the examples above that I mentioned.
