Let $X$ be a real closed field. Let us call a subset of $X$ definable if it is definable using a first-order formula in the language of ordered fields without parameters from $X$. And let us call a subset of $X$ inductive if it contains $0$ and if it contains $x$, then it contains $x+1$. And let $M$ be the intersection of all definable inductive subsets of $X$.
Then my question is, what are the properties of $M$, as a model for the language for first-order arithmetic? I don’t think $M$ is necessarily isomorphic to $\mathbb{N}$, but clearly it satisfies the axioms of Robinson arithmetic at least. But how strong a form of induction must it satisfy?
Best Answer
If $\varphi(x)$ is a formula and $M$ is a model, I will write $\varphi(M)$ for the subset of $M$ defined by $\varphi(x)$.
In any real closed field $R$, a formula with one free variable (even with parameters from $R$) defines a finite union of points of $R$ and intervals with endpoints in $R$ or $\pm \infty$ (this is called o-minimality). If $\varphi(x)$ is such a formula without parameters, then we can interpret this formula in $\mathbb{Q}^r$, the field of real algebraic numbers (this is the real closure of $\mathbb{Q}$, and the prime model of the theory of real closed fields). Since $R$ is an elementary extension of $\mathbb{Q}^r$, it follows that $\varphi(R)$ is a finite union of points in $\mathbb{Q}^r$ and intervals with endpoints in $\mathbb{Q}^r$ or $\pm \infty$.
[For example, if $\varphi(\mathbb{Q}^r) = \{0\}\cup (\sqrt{2},\infty)$, then $\mathbb{Q}^r\models \forall x(\varphi(x) \leftrightarrow (x = 0\lor \sqrt{2}<x))$, so also $R\models \forall x(\varphi(x) \leftrightarrow (x = 0\lor \sqrt{2}<x))$, and hence $\varphi(R) = \{0\}\cup (\sqrt{2},\infty)$.]
Now if $\varphi(R)$ is inductive, then $\varphi(\mathbb{Q}^r)$ is unbounded above in $\mathbb{Q}^r$. Thus there is a real algebraic number $a$ such that $\varphi(\mathbb{Q}^r)$ contains $(a,\infty)$, and hence $\varphi(R)$ also contains $(a,\infty)$. In particular, $\varphi(R)$ contains $\mathbb{N}\cup \{x\in R\mid x > \mathbb{N}\}$.
It remains to show that the intersection of all inductive definable sets (without parameters) in $R$ is exactly $\mathbb{N}\cup \{x\in R\mid x > \mathbb{N}\}$. This is just as in nombre's answer: For each natural number $n$, the set $S_n = \{0,1,\dots,n\}\cup (n,\infty)$ is definable, and $\bigcap_{n\in \mathbb{N}} S_n = \mathbb{N}\cup \{x\in R\mid x > \mathbb{N}\}$.
Note that if $R$ is Archimedean, then $\mathbb{N}\cup \{x\in R\mid x > \mathbb{N}\} = \mathbb{N}$. But if $R$ is non-Archimedean, this set is much larger, and it doesn't look much like a model of arithemetic. In particular, the infinite elements are densely ordered. (Though my earlier comment was wrong: it seems it does satisfy the axioms of Robinson artithmetic. I forgot how weak the axioms of Q are.)
Regarding induction: If $R$ is Archimedean, then $M = \mathbb{N}$ so of course it satisfies full (second-order) induction.
If $R$ is non-Archimedean, then $M$ fails induction already for the formula $\psi(x)$ which says "$x$ is odd or $x+1$ is odd": $$\forall y\,\forall z\, (y+y \neq x) \lor (z+z\neq x+1).$$
This formula is false for all infinite elements of $M$, but it is true for all finite elements, so it defines $\mathbb{N}$ in $M$. Really, you should think of $M$ as looking like a definable copy of $\mathbb{N}$ with a bunch of extraneous densely-ordered junk on top. If you're trying to find a reasonable model of arithmetic inside a real closed field, I think nombre's comment above about integer parts is much more relevant than this construction.
Quantifier-free induction: In the comments, you asked whether $M$ satisfies induction for quantifier-free formulas. The answer is yes, but this is a bit silly, since quantifier-free formulas just define finite unions of points and intervals.
Suppose $\varphi(x)$ is a quantifier-free formula. Let's allow $\varphi$ to have parameters and include the symbol $\leq$. Since $\varphi$ is quantifier-free, $\varphi(M) = \varphi(R)\cap M$. Suppose $\varphi(M)$ is inductive. Then it contains $\mathbb{N}$, so we can focus on the infinite elements of $\varphi(M)$, which are exactly the infinite elements of $\varphi(R)$. Suppose for contradiction that $\varphi(M)\neq M$. Since $\varphi(R)$ is a finite union of points and intervals with endpoints in $R$ or $\pm \infty$, and $\varphi(R)$ is unbounded in $\mathbb{N}$, there is some infinite $a\in R$ such that $(-\infty,a)\cap \{x\in R\mid x > \mathbb{N}\}$ is contained in $\varphi(R)$, but there exists $\varepsilon$ such that $0<\varepsilon < 1$, and $a+\varepsilon\notin \varphi(R)$. Then $a+\varepsilon - 1\in \varphi(R)$, contradicting inductiveness.