Proof intersection is finite and non-empty

Course: Analysis (1st year course).

Question:

If $A_3$ is a subset of $A_2$ and $A_2$ is a subset of $A_1$ and so on... are all finite, nonempty sets of real numbers, then the intersection $\bigcap_{n = 1}^{\infty} A_{n}$ is finite and non-empty.

My shot:

Proof by contradiction. Assume $\bigcap_{n = 1}^{\infty}A_n$ is empty.

Let $x$ be part of $A_{1}$. Then $x \not \in A_{k}$ for some $k>1$, because otherwise $\bigcap_{n = 1}^{\infty} A_{n}$ is non-empty. From here I get stuck.

EDIT: Somebody erroneously edited my message. The way of inclusion is opposite.

• Hint: the sizes of $A_n$ are finite positive integers, and they are non-increasing. – Steve D Apr 16 '16 at 21:41
• For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Apr 17 '16 at 11:18

By contradiction, suppose $\cap_{n\in N}A_n=\phi.$ For each $x\in A_1$ let $f(x)$ be the least $n\in N$ such that $x\not \in A_n.$ Then $B=\{f(x):x\in A_1\}$ is a finite subset of $N,$ so there exists $m\in N$ such that $\forall n\in B\;(m\geq n).$
For such an $m,$ we have $\forall x\in A_1\;(x\not \in A_m)$ because $\forall x\in A_1\;(x\not \in A_{f(x)}\supset$ $\cap_{j=f(x)}^m A_j\supset A_m).$ But then $A_m\cap A_1=\phi$ , contradicting $\phi \ne A_m \subset \cap_{j=1}^mA_j\subset A_1.$
Let $m$ be the minimum of the sizes of the nonempty finite sets $A_1,A_2,\dots$. Since they are nonempty, each $|A_k|\ge 1$, hence $m\ge 1$.
Now, $m$ is the minimum, so one of the sets $A_k$ has in effect $m$ elements. But then all $A_j$ with $j\ge k$ also has $m$ elements. So $A_k$ is repeating until the end, and this will be the intersection as well.
So I'll claim $\bigcap_{n = 1}^{\infty} A_{n} = A_{1}$, which is finite. Recall that $X \subseteq Y$ means that if $x \in X$, then $y \in Y$. So let $x \in A_{1}$, then $A_{1} \subseteq A_{2} \Rightarrow x \in A_{2}, A_{2} \subseteq A_{3} \Rightarrow x \in A_{3}$, and on, so $x \in A_{n}$ for all $n$. But if $y \not \in A_{1}$, then $y \not \in \cap_{n = 1}^{\infty} A_{n}$. So an element is in $\bigcap_{n = 1}^{\infty} A_{n}$ if and only if it is in $A_{1}$, thus $\bigcap_{n = 1}^{\infty} A_n = A_1$, and $A_1$ is finite.