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!