The Archimedean Property

Theorem 0.3 [Archimedean Principle]

$\mathbb{N}$ is unbounded. For all $M > 0$, there exists $n \in \mathbb{N}$ such that $n > M$.

Proof

Let $M > 0$. If $M < 1$, then $n = 1 > M$. Now, suppose that $M > 1$. Let

$$ E = \{n \in \mathbb{N}: n \leq M\}. $$

$E$ is a bounded set and non-empty since $1 \in E$. Then by the completeness axiom, $\sup E$ exists. Let $n_0 = \sup E$. By Lemma/Theorem 1.15, $n_0 \in E$. Then $n_0$ is a positive integer. Therefore, $n_0 + 1$ is also a positive integer. However, it is not in $E$ and thus $n_0 + 1 > M$ is our desired integer. $\ \blacksquare$

References

Lecture Notes by Professor Chun Kit Lai