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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

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    $\begingroup$ For the first point, let $A_n=[-1/n,1/n]\subseteq \Bbb R$. Then the intersection is just $\{0\}$. Even better, let $A_n=(n,\infty)$, and the intersection is empty. I think the main difference is that with infinite sets, you can have an infinite chain of strictly decreasing, non-empty sets, while that's impossible in the finite case. $\endgroup$ – Arthur Jun 1 '15 at 12:21
  • $\begingroup$ @arthur That is exactly the point I was thinking of making having read the answers but not the comments. I think it deserves a mention in some answer, because really the issue is one of the existence of a sequence. This generalizes quite a bit the easy basic generalization is assume all $A_\alpha$ are countable and you're intersecting over an uncountable set. (I'm sure full generality is just an appropriate use of cofinalities but would have to think a bit.) $\endgroup$ – DRF Jun 2 '15 at 12:19
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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.

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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.

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