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I'm working through the book Understanding Analysis by Stephen Abbott and encountered this problem:

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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.

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

2 Answers 2

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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

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Bach100
    Aug 25, 2021 at 9:00
  • $\begingroup$ @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. $\endgroup$
    – user1729
    Aug 25, 2021 at 9:22
  • $\begingroup$ 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). $\endgroup$
    – Mor A.
    Aug 25, 2021 at 10:36
  • $\begingroup$ @MorA. Whoopse - I means to intersect them! This makes it slightly messier, but still more natural than my original thought of $A_i:=\{0\}\cup[i, \infty)$. $\endgroup$
    – user1729
    Aug 25, 2021 at 11:37
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    $\begingroup$ (I've now corrected this error, and added in the simpler example.) $\endgroup$
    – user1729
    Aug 25, 2021 at 11:47

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