# Archimedean Property

Show that for every real number $y>0$, $$\bigcap_{n=1}^{\infty} (0, y/n] = \emptyset$$

So this would mean that $0< x \leq y/n$ for every positive integer $n$ which contradicts the Archimdean property?

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By "this" you mean "$x\in\bigcap_{n=1}^{\infty} (0, y/n]$". This could be written more clearly, but the answer is yes. – Jonas Meyer Jun 20 '11 at 21:44
Yes, it would mean that. Because $x\leq y/n$ holds if and only if $nx\leq y$. – Arturo Magidin Jun 20 '11 at 21:44
Very good thinking, perfectly correct. Depending on what tools you have by now, it may have been intended that you first use the "nested interval" property to show that $\bigcap[0,y/n]$ only contains the point $0$. – André Nicolas Jun 20 '11 at 21:53

Nothing wrong with the proof by contradiction, but just to be non-contrary let me give a direct proof that the archimedean property implies the intersection is empty.

Note that $\bigcap\limits_{n=1}^{\infty}(0,y/n]\subseteq (0,y]$. Therefore, $$\bigcap_{n=1}^{\infty}(0,y/n] = \left(\cap_{n=1}^{\infty}(0,y/n]\right)\cap(0,y].$$ Now let $x\in (0,y]$. By the Archimedean property, since $0\lt x$ and $0\lt y$, there exists $k\in\mathbb{N}$ such that $kx\gt y$. Therefore, $x\gt y/k$, so $x\notin (0,y/k]$. Since $\bigcap\limits_{n=1}^{\infty}(0,y/n]\subseteq (0,y/k]$, we conclude that $x\notin \bigcap\limits_{n=1}^{\infty}(0,y/n]$.

That is: for every $x$, $x\in (0,y]$ implies $x\notin \bigcap\limits_{n=1}^{\infty}(0,y/n]$. Therefore, $$\left(\bigcap_{n=1}^{\infty}(0,y/n]\right)\cap (0,y] = \emptyset.$$ Thus, $$\bigcap_{n=1}^{\infty}(0,y/n] = \left(\bigcap_{n=1}^{\infty}(0,y/n]\right)\cap(0,y] = \emptyset,$$ proving the intersection is empty.

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@Zev: Thanks for the assist. – Arturo Magidin Jun 21 '11 at 3:21
No problem! – Zev Chonoles Jun 21 '11 at 3:22
@Theo: Yes: I forgot that \cap does not automatically turn into an operator under suitable circumstances (the way that - does) and has to be manually made into one; and I navigated away instead of checking. – Arturo Magidin Jun 21 '11 at 4:05
Oh, so disappointing... I thought I could tell you didn't know about LaTeX :) – t.b. Jun 21 '11 at 4:12
@Theo: Plenty of that around, trust me; especially since I learned PlainTeX by reading The LaTeX Book (multiple times, as required) and stick to what works. Took me forever to switch to align from eqnarray and array... – Arturo Magidin Jun 21 '11 at 4:14

Indeed if $x\in\bigcap_{n=1}^\infty (0,y/n]$ then for every $n\in\mathbb N$ we have that $0<x\le y/(n+1)<y/n$.

That is for every $n\in\mathbb N$ we have $0<x<y/n$, since $\lim_{n\to\infty}y/n = 0$ we have that every positive real number is smaller than only finitely many $y/n$.

The $x$ as above does not have this property, and indeed it will be a non-Archimedean infinitesimal number.

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