To prove $\cap_{n=1}^{\infty} A_n $ is non empty

Question

To prove $$\cap_{n=1}^{\infty} A_n $$ is non empty where $A_n , n \in \mathbb{N}$ are non empty and finite sets and $A_1 \supseteq A_2 \supseteq A_3 ...$

Let this intersection be empty

Let $a \in A_1$. So $\exists k \in N , k >1$ such that $x \notin A_k $ and so $a \notin \cap_{n=k}^{\infty} A_n $.

Also $x \in A_{k-1}$ and so $x \in A_k$ which is a contradiction to fact that $x \notin A_k $

How do i proceed ? Any Hints ?

Answer 1

The finiteness of the sets is crucial, since otherwise we could, for example, let $A_n=(0,1/n)$, for which $A_1\supseteq A_2\supseteq A_3\cdots$ but $\cap_{n=1}^\infty A_n=\emptyset$.

Now if the intersection were empty, then each element $a\in A_1$ must be absent from some set $A_{n(a)}$ with $n(a)\gt1$. In that case, let $N=\sum_{a\in A_1}n(a)$, which is finite since $A_1$ is finite. Then $A_N\subseteq A_{n(a)}$ since $N\ge n(a)$, so $a\not\in A_N$ for each $a$, hence $A_N=\emptyset$. But that contradicts the assumption that the individual sets are all non-empty, hence the intersection must be non-empty.

Answer 2

Let $a \in A_1$. So $\exists k \in N , k >1$ such that $x \notin A_k $

This is false. For example, you could have $A_1=A_2=A_3=\cdots$ which shows that what you wrote is not true.

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To actually solve your problem, you need to prove that the intersection is not empty, i.e. that there exists some element in the intersection.

To do that, here are some guidelines:

1. Think about the series of numbers $|A_1|, |A_2|, |A_3|,\dots$
2. Prove that the series above is a non-increasing sequence of positive integers.
3. From $2$, conclude that the series must be constant from some point on,
4. From that, conclude that from some point on, all $A_k$ are the same set.