# 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.

• The symbol $\subseteq$ means "is a subset of", not "is a proper subset of", so it could be that for example $A_5=A_6=A_7=...$. Aug 25, 2021 at 8:25
• For a), look at the sets$[n,\infty)$. Aug 25, 2021 at 8:46
• For (a), it can be useful to compare/contrast $A_n = [-1/n,1/n] \subset\Bbb{R}$ with $A_n = (-1/n,1/n) \subset \Bbb{R}$. Oct 18, 2021 at 17:07

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

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.

• Can you help me understand the intuition behind this? Yes we can find counter-examples that invalidate it but what is the general logic behind all of this? Aug 25, 2021 at 9:00
• @Bach100 in the two examples here we have $A_i=B_i\cup\{0\}$. This always happens: if $\cap S_i$ is finite then there is a fixed set $X$ and a sequence $(T_i)$ such that $S_i=T_i\cup X$ and $\cap T_i$ is empty. So your intuition that the sets should be empty is almost correct. Aug 25, 2021 at 9:22
• You do not have $B_1\supseteq B_2\supseteq B_3 \supseteq \cdots$ (for example $3\in B_3\setminus B_2$) so this is not a counterexample for (a). Aug 25, 2021 at 10:36
• @MorA. Whoopse - I means to intersect them! This makes it slightly messier, but still more natural than my original thought of $A_i:=\{0\}\cup[i, \infty)$. Aug 25, 2021 at 11:37
• (I've now corrected this error, and added in the simpler example.) Aug 25, 2021 at 11:47