Convergence of a nested sequence of sets that all contain a common set

Question

I spent hours looking for answers to my question, but I could not find anything.

I am looking for a proof that a nested sequence of sets $A_{n+1} \subseteq A_n \subseteq ... \subseteq A_0$ that all contain a nonempty set B ($B \subseteq A_n, \forall n$) is converging towards a non-empty set $A_{\infty} $

The sets $A_n$ are linear convex sets of $ \mathbb R^p $. So I guess I can say they are closed (?)

The problem is that they are not bounded. So, I cannot use Cantor's intersection theorem.

I know there are seome counter examples of nested closed unbounded sets that do not converge towards a nonempty set, such as $F_n = [n,\infty[$.

But in my case, I have the additional condition that $B \subseteq A_n, \forall n$ which should help.

Anyone knows the theorem saying that my sequence is converging?

Thanks a lot!

Quentin

Answer 1

So I'm not really sure what you mean by "converging" here, as that's not frequently a term that's used for sets (although you could use it once you define a topology on the space of all subsets of a set). In particular, although the Cantor Intersection Theorem does use topological concepts (i.e. compactness), it doesn't use the idea of "convergence" of sequences of sets but only talks about their intersection.

That said, it's a simple fact about sets that

$$ B \subset A_i \forall i \implies B \subset \cap_{i=o}^{\infty} A_i, $$ and that's true of any sequences of sets without even having to talk about topology. So if you just wanted to show that the intersection of all the $A_i$ is not empty then you could define $A_{\infty}$ to be $\cap_{i=o}^{\infty} A_i$ and then say that the $A_i$ "converge to" it, (and the existence of a set $B$ like you described guarantees that $\cap_{i=o}^{\infty} A_i$ is not empty). But really that wouldn't be very good use of mathematical language.

If I were writing up a problem and all I cared about is that the intersection isn't empty, I would write "Because all of the $A_i$ contain the (non-empty) set $B$, we can see that $\cap_{i=o}^{\infty} A_i$ is non-empty." and I think that would satisfy most people. If it's not clear to you why that's true then add a comment and we can dig into more details.

Answer 2

So, if I understand well your comments, the infinite intersection of a sequence of sets always exists. Am I right ?

Therefore, for a nested sequence of sets $A_{n+1} ⊆ A_n ⊆...⊆ A_0 $, "convergence", i.e. "$\lim_{n\to 0} A_n$", always exists ?

Answer 3

What I mean is that, contrary to the nested sequence I am refering to, the sequence of sets $ A_n = [0,1+(-1)^n] $ does not "converge", because it is always "jumping" between the intervals $ \{ 0 \} $ and $[0,2]$.