The set N of natural numbers is unbounded above in R.
If N were bounded above, then by the completeness axiom it would have a least upper bound, say sup N = m. Since m is a least upper bound, m – 1 is not an upper bound of N. Thus there exists an n in N such that n > m – 1. But then n + 1 > m , and since n + 1 ∈ N, this contradicts m being an upper bound of N . ♦
(from "Analysis with Introduction to Proof" 5th edition)