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? 
 A: 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.
A: 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.
