Difference between intersection of infinite sets having finite, and having infinite elements

Question

I could find individual answers for both of these, but can someone compare how being a finite set or an infinite changes the final outcome?

(a) If A1 ⊇ A2 ⊇ A3 ⊇ A4 · · · are all sets containing an infinite number of elements, then the intersection $$ \bigcap_{n=1}^{\infty} A_n $$ is infinite as well. - False

(b) If A1 ⊇ A2 ⊇ A3 ⊇ A4 · · · are all finite, nonempty sets of real numbers, then the intersection $$ \bigcap_{n=1}^{\infty} A_n $$ is finite and nonempty. - True

This is from the book Stephen Abbott, Understanding Analysis

Answer 1

I think you're asking what the relationship is between the two. The relationship is clearest when the statements are changed slightly.

Let $A_1 \supseteq A_2 \supseteq A_3 \supseteq \dots$ be a descending sequence of nonempty sets. Must their intersection be nonempty?

The answer is no in general, but yes if all the sets (or even just one of the sets) are assumed finite. This appears paradoxical because finite sets are smaller than infinite ones, so one would naïvely expect their intersection to be smaller too.

Answer 2

In (b) the sequence is bound to stabilize: some $n$ exists such that $k\geq n\implies A_k=A_n$.

The finite number of elements of $A_1$ can only be 'diminished' a finite number of times. This combined with the condition that $A_n\neq\varnothing$ assures that the intersection is not empty.

In (a) that obstacle does not appear. Every element in $A_1$ can be "thrown out" at some time.