Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
@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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.