Define the set of real numbers $\mathbb{R}$ by means of the following axiom: There exists a totally ordered field $(\mathbb{R},+,\cdot,\leq)$ which is Dedekind complete. We also assume that $a\leq b$ implies $a+c\leq b+c$ for all $a,b,c\in \mathbb{R}$ and $a\leq b$, $0\leq c$ implies $ac\leq bc$ for all $a,b,c\in \mathbb{R}$.
We derive the well-ordering principle for $\mathbb{N}$. Namely, we claim that every non-empty subset $S$ of $\mathbb{N}$ has a minimum.
We prove this fact as follows:
Since $\mathbb{N}$ is bounded below, the completeness of $\mathbb{R}$ provides us with $\alpha:=\inf S$. It suffices to show that $\alpha\in S$.
Arguing by contradiction, suppose that $\alpha\notin S$. Then $\alpha+1/2$ is not a lower bound; that is, we can find $m\in S$ so that $\alpha<m<\alpha+1/2$. Again, since $m$ is not a lower bound, one can find $m^\prime\in S$ such that $\alpha<m^\prime<m<\alpha+1/2$. This means that $m-m^\prime<\frac{1}{2}$. But this is impossible, as the difference of every pair of different natural numbers cannot be lower that one.
My question is: how can one derive from the definition of real numbers, the fact: for all $m,m^\prime\in\mathbb{N}$, $m>m^\prime$ one has $m-m^\prime\geq 1$? I would do by induction, however, as far as I know, the well-ordering principle is equivalent to the induction principle, so it would be tricky to use induction to derive the well-ordering principle. Should one to include the induction principle as an axiom?
