How can there be an infinite number of nesting finite sets? I'm working through the book Understanding Analysis by Stephen Abbott and encountered this problem:

And I can't really wrap my head around how there could be an infinite series of nesting sets each containing finite elements? And how their intersection will be non-empty? Since they have finite elements wouldn't it mean that at one point they would 'run out' of elements and become an empty set?
I would really appreciate if someone could also explain to me the logic behind why (a) is false. I'm self studying Analysis and everything has been really confusing.
 A: Notice that you don't need them to be different. Just take all sets to be equal and you're done. In fact the only solution is that after some $i$ all $A_i$'s will be equal
A: An easy counter-example to (a) is the sequence of sets $(A_i)_{i\geq0}$ where $$A_i:=[i, \infty)\cup\{0\}.$$
Here, $\cap A_i=\{0\}$ which is non-empty.
Of course, we kind of cheated here: we added in the intersection manually, as $\cap[i, \infty)=\emptyset$. However, this always happens: if $\cap S_i$ is finite then there is a fixed finite set $X$ and a sequence $(T_i)$ such that $S_i=T_i\cup X$ and $\cap T_i$ is empty (in fact, $X=\cap S_i$). So your intuition that the sets should be empty is almost correct.

A more involved, but also more natural, counter-example for (a) is the sequence of sets $(A_i)_{i\geq0}$ where $$A_i:=\{in\mid n\in\mathbb{N}\cup\{0\}\}\cap A_{i-1}.$$
This is a chain of infinite sets with $\cap A_i=\{0\}$:

*

*Intersecting with previous terms ensures that this is indeed a chain of sets.

*Each $A_i$ is infinite, as if $x\in A_{i-1}$ then $\operatorname{lcm}(x, i)n\in A_i$ for all $n\in\mathbb{N}$.

*The intersection is $\{0\}$ as $0\in A_i$ for all $i$, while if $x\neq0$ is such that $x\in\cap A_i$ then $x$ is contained in every $A_i$, and so $i$ divides $x$ for all $i\in\mathbb{N}\cup\{0\}$, but of course $i$ can only divide $x$ if it is smaller than $x$. Hence, there can be no non-zero $x\in \cap A_i$ as required.

Note that taking the sequence of sets $(B_i)_{i\geq0}$ where $B_i:=\{in\mid n\in\mathbb{N}_{>0}\}\cap B_{i-1}$, gives $\cap B_i=\emptyset$. The only difference here is that we have removed $0$ from each set.
