4
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ If you have a sufficient condition for a subspace to be relatively compact can't you just take the same condition+closed as a sufficient condition for compactness? $\endgroup$ – Alessandro Codenotti Oct 17 at 21:52
  • $\begingroup$ @AlessandroCodenotti yes you are right. Relatively compact means closure is compact, so if we add closed, the closure is the set itself, so we are done. Right? Can you think of an example of such set? That is, closed, bounded and $$\lim_{N\rightarrow\infty}\sup_{(x_{j})\in K}\sum_{j=N+1}^{\infty}|x_{j}|=0?$$ $\endgroup$ – JacobsonRadical Oct 17 at 22:32
  • $\begingroup$ The use of the word 'subspace' is confusing. The only linear subspace that is compact is $\{0\}$. $\endgroup$ – Kavi Rama Murthy Oct 17 at 23:35
  • $\begingroup$ @KaviRamaMurthy sorry. I mean subset. Post edited. $\endgroup$ – JacobsonRadical Oct 18 at 0:40
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Really nice answer. Thank you! $\endgroup$ – JacobsonRadical Oct 19 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.