Where is the copy of $\mathbb{N}$ in the constructible hierarchy relative to a real closed field?

Question

Let $X$ be a real closed field. Let us define a constructible hierarchy relative to $X$ is defined as follows. (This is slightly nonstandard terminology.). Let $L_0(X)=X$. For any ordinal $\beta$, let $L_{\beta+1}(X)=Def(L_{\beta+1}(X))$. For any limit ordinal $\gamma$, let $L_\gamma(X)=\cup_{\beta<\gamma}L_\beta$. And finally let $L(X) = \cup_\alpha L_\alpha$.

Now let $M=\{n1_X: n\in\mathbb{N}\}$. Then my question is, what is the smallest ordinal $\alpha$ such that $M$ is guaranteed to be an element of $L_{\alpha}(X)$?

Or is it consistent that $M\notin L(X)$? What if we were to add the axiom $V=L(X)$?

Answer 1

To avoid notational clash, I'll use the notation $D_\alpha(X)$ to describe the hierarchy built on an RCF $X=(A;f_1,f_2)$ with underlying set $A$, addition function $f_1$, and multiplication function $f_2$, defined precisely as follows:

• At the successor and limit steps we take definable powersets and unions respectively, as usual.

• We start with $D_0(X)=A\cup A^2\cup\{f_1,f_2\}$.

Here are a couple quick comments to demonstrate that $D_0(X)$ really does have at least the "bare minimum" of expressive power we want for a set-theoretic implementation of an RCF:

• We have $f_1, f_2\subseteq D_0(X)$ (and consequently $f_1,f_2$ are definable subsets of $D_0(X)$ since also $f_1,f_2\in D_0(X)$). This is because $A^2\subseteq D_0(X)$ and $f_1,f_2\subseteq A^2$.

• We have that $A$ is a definable subset of $D_0(X)$ - e.g. as "The set of left coordinates of elements of $f_1$."

• We can tell which of $f_1$ and $f_2$ is addition and which is multiplication, by asking which has an annihilator.

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Now right away, we can make the following observation. As we go along the $D$-hierarchy, we "accidentally" wind up following the usual construction of $L$. In particular, we have $A^{<\omega}\subseteq D_\omega(X)$. This lets us implement the "natural" definition of $M$ in $D_{\omega+1}(X)$: "$M$ is the set of $m\in A$ such that there is some finite sequence of elements of $A$ whose first term is $1_X$, whose last term is $m$, and whose $(i+1)$th term is the $i$th term $+_X1_X$." This gives us the following:

$\alpha\le\omega+1.$

Can we do better? Well, at least for some presentations we can easily. Specifically, suppose that $$\mathcal{P}(A)\cap D_1(X)=Def(X),$$ where $Def(X)$ is the set of subsets of $A$ which are definable in the RCF $X$ in the model-theoretic sense. Then by o-minimality of RCF, we have that the following are equivalent for $U\in \mathcal{P}(A)\cap D_1(X)=Def(X)$:

• $U$ is discrete, has $1_X$ as its least element, and for each $d\in U$ with $d\not=1_X$ we have $d-_X1_X\in U$.

• $U=\{1\cdot 1_X, 2\cdot 1_X, ..., n\cdot 1_X\}$ for some $n\in\mathbb{N}_{\ge 1}$.

This gives us $M\in D_2(X)$: we have $m\in M$ iff there is some $U\in \mathcal{P}(A)\cap D_1(X)=Def(X)$ satisfying the above two bulletpoints with $m\in U$. Consequently, we have:

Restricted to "model-theoretically efficient" presentations of RCFs, that is, ones where $D_1(X)$ is "minimal," we have $\alpha=2$.

(It's easy to show $\alpha>1$.)

Moreover, we can get this unconditionally if $X$ is additionally Archimedean. This is because we can simply add the criterion that $U$ be bounded above and below; the only subsets of $A$ which are bounded above and below, contain $1_X$, and are closed under subtracting $1_X$ from ever non-$1_X$ element are the sets of the form $\{1\cdot 1_X, 2\cdot 1_X, ..., n\cdot 1_X\}$ for some $n\in\mathbb{N}$. That is:

If $X$ is Archimedean, then $\alpha=2$.

However, we run into a problem if $X$ is non-Archimedean and is presented in such a way that non-definable-in-$X$ subsets of $A$ show up in $D_1(X)$. In general I don't see a way to improve on the $\omega+1$ bound.

Conjecture: There is an RCF $X$ whose $\alpha$ is exactly $\omega+1$.