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$\bigcap I_i=\emptyset$ where $I_i=(0, \frac{1}{i})$

What does it mean that the infinite intersection is empty?

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  • $\begingroup$ It means there exists no point in all of these sets $I_i$? $\endgroup$ – John Oct 17 '14 at 16:01
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    $\begingroup$ See also math.stackexchange.com/a/228525/589. $\endgroup$ – lhf Oct 17 '14 at 17:13
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This means there cannot be a real number $x$ such that $0 < x < \dfrac{1}{n}$, for all $n$. For if there were such an $x$, then $x \leq \displaystyle \lim_{n \to \infty} \dfrac{1}{n} = 0$. Thus $x \leq 0$, and $x > 0$ , a contradiction.

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    $\begingroup$ Although perfectly valid, I think that one should note that we typically talk about such things before we formalize concepts such as limits, and that the original statement that there does not exist a real number satisfying $0 < x < \frac{1}{n}$ for all positive integers $n$ is equivalent to the Archimedean property. $\endgroup$ – heropup Oct 17 '14 at 17:06

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