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I've seen this problem before, but can't remember how to finish it:

Define an indexed family of sets $ \{A_i : i \in \mathbb{N} \}$ in which for any $m,n\in \mathbb{N}, A_m \cap A_n \not= \emptyset$ and $\bigcap A_i = \emptyset$. The closest I came was something to the effect of $A_i = \{(0, 1/n): n \in \mathbb{N} \}$, but I know that doesn't meet the last criterion.

Suggestions on how to fix/finish it? Am I even as close as I think I am?

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    $\begingroup$ To the best of my knowledge, you are done. $\endgroup$
    – Lord_Farin
    Apr 8, 2013 at 21:28
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    $\begingroup$ Why do you think it doesn't meet the last criterion? It does. $\endgroup$
    – Lord Soth
    Apr 8, 2013 at 21:29
  • $\begingroup$ Cool--I guess I thought there would always be an overlap at the end; is that eliminated because there's always a smaller interval than any nth interval? $\endgroup$
    – js14
    Apr 8, 2013 at 21:34
  • $\begingroup$ You may go like this: Suppose $a>0$ is a member of the infinite intersection. Then, find an index $n$ such that $a\notin A_n$ (choosing $n$ such that $a > \frac{1}{n}$ works) that leads to a contradiction. The fact that $0$ is not a member of the intersection is obvious. $\endgroup$
    – Lord Soth
    Apr 8, 2013 at 21:39

3 Answers 3

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Further to Hagen von Eitzen and Asaf Karagila's answers, you could set $$A_1=\{0,1\}, \quad A_2 = \{ 0,2 \}, \quad A_n=\{1,2\}\ \text{for}\ n \ge 3$$ No need for any clever reasoning about infinite sets.

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There are several obvious ways to do this, and some less obvious ways too:

  • $A_n=\{x\in\Bbb R\mid n<x\}$;
  • $A_n=\left(0,\frac1n\right)$;
  • $A_n=\Bbb N\setminus\{n\}$;
  • $A_n=\{k\in\Bbb N\mid k\text{ is positive and divisible by the }n\text{-th prime number}\}$;
  • $A_n=\{n\cdot k\mid k\in\Bbb N\setminus\{0\}\}$ for $n>0$, and $A_0=\Bbb N$ (if $x\in A_n$ for all $n$ then $x$ can be divided by every number, which is impossible).

There are many more ways which are interesting and can reflect on many mathematical properties related to finiteness and countability.

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How about $A_i =\{n\in\mathbb N\mid n>i\}$? But your choice, more properly written $A_i=(0,1/i)$, is also fine.

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  • $\begingroup$ Great--thanks so much! $\endgroup$
    – js14
    Apr 8, 2013 at 21:35

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