If $ ... A_3 \subseteq A_2 \subseteq A_1$ are all finite, nonempty sets of real numbers, then $ \bigcap_{i=1}^{\infty}A_{i} $ is finite & nonempty First, I noted that, $|A_1| \geq |A_2| \geq ... \geq 1, |A_i| \in \mathbb{N}$ Let $$A(x) := \{ i : |A_i| = x \}$$ $$B:=\{|A_i| : i \in \mathbb{N} \}$$
Then B is bounded below by 1. By the well ordering principle, every non empty set of positive integers has a least element. Let $j$ be the least element of B. Then $$A(j) := \{i : |A_i|=j \}$$Then since A(j) is also a set of positive integers, it also has a least element, let's call it $i$. Then $A_i$ is the first occurrence of the smallest sized set. Note $A_i$ is non empty and finite. Then $\bigcap_{j=1}^{i} A_j = i$. Because all following sets are subsets of $A_i$, and because $A_i$ is the smallest sized subset, it follows that every subset $Aj, j > i$ is equal to $A_i$ in order to remain contained in $A_i$. Therefore, $\bigcap_{j=1}^{\infty} A_j = \bigcap_{j=1}^{i} A_j \cap \bigcap_{j=i+1}^{\infty}A_i = i$
Is my proof correct? Also, is there a way to prove this without using the well ordering principle? My textbook didn't mention it, which is why I ask.
 A: Your proof is correct. An alternative approach would be via the following observation:
Claim: The set $S = \{i\in \mathbb{N}:\, A_{i+1}\neq A_i\}$ is finite.
Suppose for the sake of contradiction that $S$ is not finite. In this case we can find a sequence of indices $i_1<i_2<\dots<i_n<i_{n+1}<\dots$ such that
$$A_{i_j+1}\subseteq A_{i_j}\quad \text{but}\quad A_{i_j+1}\neq A_{i_j},$$
for all $j$. Then, by picking an element $x_j\in A_{i_j}\setminus A_{i_j+1}$ we end up with pairwise distinct elements $(x_j)_{j=1}^{\infty}$ all contained in $A_1$. This means that $A_1$ must be an infinite set which leads to the desired contradiction.
Now, since $S$ is finite there would exist an element $s^*=\max(S)$. Then, for every $k\geq s^*$ by the fact that $s^*$ is the maximum of $S$ we have that $A_{s^*}=A_k$ for all $k\geq s^*$. This is equivalent to
$$\bigcap_{n=1}^{\infty}A_n = A_{s^*}$$
Hence, the intersection is non empty and finite since it is equal to $A_{s^*}$.
A: Your proof is correct. More simply: the sequence of natural numbers $|A_n|$ is monotonically decreasing hence eventually constant:
$$\exists N\quad\forall n\ge N\quad|A_n|=|A_N|.$$
For every $n\ge N,$ since moreover $A_n\subseteq A_N,$ we deduce $A_n=A_N.$
Therefore,
$$\bigcap_{n\in\Bbb N}A_n=A_N.$$
A: To better hide the well-ordering principle, look at the finite set $A_1=\{x_1,\ldots,x_m\}$.
If we assume that the intersection is empty, the there exit naturals $n_1,\ldots,n_m$ such that $x_k\notin A_{n_k}$ — and by the descending condition, also $x_k\notin A_n$ for any $n\ge n_k$.
Let $N=\max\{n_1,\ldots,n_m\}$. As $N\ge n_k$, we also have $x_k\notin A_N$. Then $A_N$ is a subset of $A_1$ that does not obtain any of the $x_k$, i.e., $A_N$ is empty — contradiction. We conclude that the intersection is not empty.
A: Because $A_1$ is finite, it has finitely many subsets,
all themselves finite.  So
$\mathscr{S} := \{A_i : i \in \mathbb{N}\}$ is finite.  Because
$\mathscr{S}$ is finite and totally ordered by inclusion, it has a
least element, $B.$ Therefore
$$
\bigcap_{i=1}^\infty A_i = \bigcap\mathscr{S} = B \ne \emptyset.
$$
