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

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

• 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. – Arthur Jun 1 '15 at 12:21
• @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.) – DRF Jun 2 '15 at 12:19

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