The intersection of an infinite descending chain of non-empty sets

Question

I am trying to prove something by contradiction and I am stuck as described below:

From the (false) assumption, I have shown that for any $i \in \mathbf{N}$, $A_i \neq \emptyset$ and that $A_1 \supseteq A_2 \supseteq A_3 \supseteq ...$ (the inclusion is not necessarily proper). Each $A_i$ is a set of functions from $\mathbf{N}$ to $\lbrace 0,1 \rbrace$. (So I believe it's not bounded or even continuous. Edited from here: Any element in each A_i is some function $f: \mathbb{N} \rightarrow \lbrace 0,1 \rbrace $ with some special property associated with $i$. So each $A_i$ is already defined. )

To achieve a contradiction useful for my proof, I either want to show that this descending chain is impossible or that $ \bigcap_{n=1}^{\infty}A_{n} \neq \emptyset$ because I can then proceed my argument.

The latter seemed obvious to me at first, but when trying to write down a proof, I couldn't. Do I need to use any special tool for this, e.g., AC? Or I can't just say either of the two?

Thanks so much!

Answer 1

The set of functions from $\Bbb{N}$ is $\{0,1\}$ is equinumerous with $\Bbb{R}$. So your problem is equivalent to the same question with sets of reals. But then it's easy to find a chain of the type you don't want: just let $A_i=(i,\infty)$.

Answer 2

Note that under certain assumptions, e.g. Cantor's theorem (about intersection of decreasing sets in a compact metric space), the intersection is guaranteed to be non-empty.

Since you mentioned continuity and boundedness, note that there is a natural topology on $2^\Bbb N$, and it is compact (in fact it is the Cantor space). So it might be reasonable to try and prove that your $A_i$ are closed sets there, in which case Cantor's theorem indeed assures that the intersection is non-empty.

If, on the other hand, the definition of the $A_i$ is too "wildly" then you can easily concoct a sequence whose intersection is empty. It depends on your sets (i.e., on how you define them).

Both Cantor's theorem, as well the existence of a sequence of sets whose intersection is empty, do not depend on the axiom of choice and are true without it.