# Nested interval property and density of rational numbers

My lecturer asks us to prove this: $$\bigcap_{n \in \mathbb{N}}\left(0,\frac{1}{n}\right)=\emptyset$$ But it seems I can find a way to prove this is not true. According to the density of rational numbers in real numbers , there exists a rational in interval $(a,b)$ . So no matter how small an interval is , we can find a number. So it is not empty. Can anyone guide me on this ? I'm very confusing now.

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If you wish to prove that the set is non-empty, you need to find a single number, $a$, that is in every interval $(0,1/n)$. Just finding possibly different numbers (rational or not) in these intervals does not accomplish this. – David Mitra Jan 19 '13 at 17:36
I see. So can I reasoning like this: By the density of rational numbers in real number , there exists a rational number in an interval. But there exists an even smaller interval which does not contain the rational number. – Idonknow Jan 19 '13 at 17:40

The density of the rationals means that we can always find a rational $x_{a,b}\in (a,b)$. The thing to note here is that $x_{a,b}$ depends on $a,b$!. More rigorously: $$\forall a,b\in \mathbb{R}:a<b\ \exists x\in \mathbb{Q}: x\in (a,b)$$ This is not the same as $$\exists x\in \mathbb{Q}\ \forall a,b\in \mathbb{R}:a<b\ x\in (a,b)$$ which you are implying. If we try to apply the density of rationals in this case we have that: $$\forall n\in \mathbb{N}\ \exists x_n\in \mathbb{Q}: x_n\in (0,\frac 1n)$$ This doesn't mean however that $$x_n\in \bigcap_{n=1}^{\infty}(0,\frac 1n)$$ as $x_n$ only needs to be a member of each own $(0,\frac 1n)$ and not for the smaller intervals.

Instead our only hope would be for this to hold for the limit. But there is a problem, the limit is $0$! Indeed as $0<x_n<\frac 1n$ $x_n$ converges and to $0$. But obviously, $$0\notin \bigcap_{n=1}^{\infty}(0,\frac 1n)$$ and so our argument collapses. In fact, $$\bigcap_{n=1}^{\infty}(0,\frac 1n)=\emptyset$$ because if it weren't then, $$\exists x\in \bigcap_{n=1}^{\infty}(0,\frac 1n)\iff 0<x<\frac 1n\ \forall n\in \mathbb{N}\iff n<\frac 1x\ \forall n\in \mathbb{N}$$ which is impossible (Archimedean Property of $\mathbb{R}$)

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so it is like an infinite loop ? I have a rational number in an interval $(0,\frac{1}{n})$ but I can find smaller interval which does not contain this rational number. – Idonknow Jan 19 '13 at 18:33
@Idonknow Pretty much. If you have intervals of the form $[0,\frac 1n)$ then such a rational number would exist ($0$) – Nameless Jan 19 '13 at 18:35
In your explanation, you talk about limit is $0$. But isnt't this mean that the sequence aproaches 0 but not equal to $0$ ? Then we can find a rational number in the interval such that it lies in the intersection ? – Idonknow Jan 20 '13 at 6:24
@Idonknow The sequence certainly "aproaches" $0$ (the limit is $0$) and is never equal to $0$ ($x_n\in (0,\frac 1n)\implies x_n>0$). If the intersection were finite such a rational would exist. But because the intersection is (countably) infinite, there is only hope for the limit to be there – Nameless Jan 20 '13 at 7:59
Actually what is $\bigcap_{n \in \mathbb{N}}{(0,\frac{1}{n}})$ ? What is the resulting set ? – Idonknow Jan 20 '13 at 8:06

Suppose that the intersection of all these intervals is nonempty. Now let $x$ belong to the intersection. It follows that $x$ is less than $1/n$ for all integers n and greater than $0$. This contradicts the Archemidean property of the real numbers.

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