Well-Ordering Principle/Proof by Restriction of Real Numbers

Proof
Let $S$ be a non-empty subset of the set of natural numbers $\N$.

We take as axiomatic that $\N$ is itself a subset of the set of real numbers $\R$.

Thus $S \subseteq \R$.

By definition:
 * $\forall n \in \N: n \ge 0$

and so:
 * $\forall n \in S: n \ge 0$

Hence $0$ is a lower bound of $S$.

This establishes the fact that $S$ is bounded below.

By the Continuum Property, we have that $S$ admits an infimum.

Hence let $b = \inf S$.

Because $b$ is the infimum of $S$, it follows that $b + 1$ is not a lower bound of $S$.

So, for some $n \in S$:
 * $n < b + 1$

$n$ is not the smallest element of $S$.

Then there exists $m \in S$ such that:


 * $b \le m < n < b + 1$

from which it follows that:
 * $0 < n - m < 1$

But there exist no natural numbers $k$ such that $0 < k < 1$.

Hence $n = b$ is the smallest element of $S$.