Is there a good metric under which a sequence of compact sets can converge to an infinite dimensional set?

Question

I have a sequence of finite-dimensional, compact sets in $L^2(\Omega)$, where $\Omega\subset \mathbf{R}^2$ is closed and bounded. The dimension grows monotonically with the sequence, and there is no assumption that the sets are nested or anything. I would like to say something about the "limiting set", but sequence doesn't converge under standard "set metrics" like the Hausdorff metric (for example, in the Hausdorff psuedometric, a limit of compact sets, if it exists, is compact, so long as the underlying metric space is complete).

Does anyone know a way to talk about convergence of finite-dimensional compact sets to an infinite-dimensional set?

P.S. This is my first post. I read the rules, but my apologies in advance for any mistakes.

Answer 1

As a general idea, without any other information on a sequence, it seems rather unlikely to find a natural metric in which it converges. Yet a metric in which it has a convergent subsequence is a more reasonable task; then you may prove that your sequence actually converges there, by means of additional informations on the particular sequence.

Here, if these compact sets $C_n$ are included in a closed ball $B$ of $L^2(\Omega)$, you may use the Hausdorff distance induced from the weak topology, which makes $B$ a metric compact space. So $\mathcal{H}(B)$ is a compact metric space, and your sequence has a convergent subsequence there, that converges to a weakly compact set. Otherwise you may just consider the trace on each ball, $C_n\cap \bar B( 0,r)$, and get by a standard diagonal argument a subsequence $C_ {n _ k}$ that converges on the Hausdorff distance of the weak topology on bounded sets.

Answer 2

If you do not have a limit, but want to talk about it you say ultralimit.

Let $(K_n)$ be your sequence of compact sets (in any complete metric space not nesessury $L^2(\Omega)$). Consider sequence of functions $$f_n=\mathop{\rm dist}\nolimits_{K_n}.$$ It is a sequence of 1-Lipschtz functions, so you can pass to its ultralimit $f_\omega$ for a fixed in advance ultrafilter $\omega$. The zero set $K_\omega$ of $f_\omega$ can be considered as the ultralimit of $K_n$.

Answer 3

Not sure if it can help, but one natural object to look at, when you have some sequence of sets (whatever the nature) is to look at the upper and lower limit.

For example, if the sequence of sets is $\{A_k\}_{k=1}^{\infty}$ you can consider:

1. The limsup of $A_n$ which is defined as follows: $$ \lim\sup A_n = \bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k, $$ which corresponds to the points who appear infinitely often in the sequence.
2. The liminf of $A_n$ which is defined as follows: $$ \lim\inf A_n = \bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty} A_k $$ which corresponds to all the points which belong to all the sets starting at least from some number.

These objects give in some sense are the natural approximations of lower and upper limit of your sequence. They also attain a lot of properties with probability, if you need it (you can look at any book in probability).

And yes, previous comments where correct that infinite-dimensional object ofcourse can be compact. You have to precise a bit better what you want to look at.

Answer 4

Ok. Suppose $\mathcal{F}_n \subset L^2(\Omega)$ such that $\mathcal{F}_n$ is spanned by $n$ linearly independent "vectors" with bounded norm, which means $L^2(\Omega)$ functions such that $||f||\leq C$ for all $f \in \mathcal{F}_n$. Then being finite-dimensional subspaces these sets will be compact in the strong topology.

However, in general, the limit will almost never be compact, since in this case it will recover a ball in $L^2(\Omega)$ which is not compact in the strong topology. You are going to need to make some more assumptions on the relationship between the sets to recover some information on the limit.