real analysis - How does the well-orderedness of the set of natural numbers follow assuming the inductive-set definition of natural numbers

Assume that $\mathbb R$ is an ordered field (i.e. $\mathbb R$ is a model of real numbers). We define the set of natural numbers $\mathbb N$ as the smallest inductive set containing $1_\mathbb R$ (multiplicative identity of the field $\mathbb R$), where by definition a set $X\subset \mathbb R$ is inductive if $x\in X$ implies $x+1_\mathbb R\in X$.

Now I wish to prove that every nonempty subset $M$ of $\mathbb N$ contains a minimal element. My book proves it as follows: If $1_\mathbb R\in M$ then $1_\mathbb R$ is the minimal element, otherwise consider the set $E:=\mathbb N - M$, which contains $1_\mathbb R$. The set $E$ must contain a natural number $n$ such that all natural numbers not larger than $n$ belong to $E$ but $n+1$ belongs to $M$; if there were no such $n$, the set $E$ which contains $1_\mathbb R$ would contain along with each of its elements $n$, the number $n+1_\mathbb R$ too, hence it would equal the whole $\mathbb N$, a contradiction. The number $n+1_\mathbb R$ so found is the minimal element of $M$.

But I do not think the bold-faced argument is correct (why $E$ would contain $n+1$ if $n\in E$?), or if it is correct according to what axioms is it correct?

Answer

Assume that there is no such $n$ as above, so that for every $n$ either for some $k\leq n$ we have $k$ is not an element of $E$ or $n+1\in E$. I prove the former case cannot happen: For $1$ it does not happen. If for $n$ it does not happen, for $n+1$ the only case to check is that $n+1$ itself is an element of $E$ as well (since all other naturals less than or equal to $n$ by the induction hypothesis is already in $E$. If $n+1$ were not in $E$ then it would be in $M$, then there would be indeed such an $n$ as in the question. So $n+1$ would be in E. Hence $E=\mathbb N$.