Archimedean Proof? I've been struggling with a concept concerning the Archimedean property proof.  That is showing my contradiction that For all $x$ in the reals, there exists $n$ in the naturals such that $n>x$.  
Okay so we assume that the naturals is bounded above and show a contradiction.  
If the naturals is bounded above, then it has a least upper bound (supremum) say $u$ 
Now consider $u-1$.  Since $u=\sup(\mathbb N)$ , $u-1$ is an element of $\mathbb N$.  (here is my first hiccup, not entirely sure why we can say $u-1$ is in $\mathbb N$) 
This implies (again not confident with this implication) that there exists a $m$ in $\mathbb N$ such that $m>u-1$.  A little bit of algebra leads to $m+1>u$. 
$m+1$ is in $\mathbb N$ and $m+1>u=\sup(\mathbb N)$ thus we have a contradiction.
Can anyone help clear up these implications that I'm not really comfortable with?  Thanks! 

I dont think you need the fact that $u\in N$(the fact is even not true) .And for the second difficulty the fact follows from the supremum property.As $u-1$ is not an upper bound so there exists a natural number greater than it.
