Math Help - Real analysis proof

1. Real analysis proof

let 0<x show that there is a unique m in N such that m-1 < or = x < m.
the book says to consider n in N : x < n and that N is well ordered

2. Originally Posted by mtlchris
let 0<x show that there is a unique m in N such that m-1 < or = x < m.
the book says to consider n in N : x < n and that N is well ordered
For $x\in\mathbb{R}$, consider the set $S_x=\{n\in\mathbb{N}:n>x\}$. By the well ordering principle, this set has a unique smallest element. Call this element $m$.

$x because $m\in S_x$. $m-1\leq x$ because if it isn't, then $m-1\in S_x$, contradicting the fact that $m=\min(S_x)$.