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It is well known that the first-order theory of any real-closed field is complete, and consequently not capable of interpreting the majority of modern mathematics.

Is this still true for the second-order theory of a real-closed field? What about third/higher order theories?

I ask because moving to second-order theories can change things, like how second-order PA is categorical while first-order PA isn’t, and because intuitively it feels like allowing enough stages of ‘higher order theory’ over the base real-closed field (think ‘inaccessibly higher order theory’) should eventually just be set theory with the elements of the field as urelements, in which case we obviously have an incomplete and robust theory.

Another intuitive reason incompleteness may arise is definability of the naturals as a subset of any real-closed field, once we have second/higher order parameters available. Any pointers are appreciated.

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  • $\begingroup$ Based on the argument here it looks like this question is sensitive to what we mean by ‘second-order theory of a real closed field’, since the Dedekind completeness axiom forces the theory to be categorical and we can define the naturals, hence interpret PA and get incompleteness. Perhaps this question would fit better over at MSE? $\endgroup$
    – Alec Rhea
    Commented Oct 19, 2023 at 2:09
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    $\begingroup$ When you say second-order theory, do you mean that you can quantify over sets or that you can quantify over relations of arbitrary arity? It's possible that this won't matter, but if you can quantify over set of arbitrary artiy, I'm fairly certain you can define the natural numbers and do all of second-order arithmetic. $\endgroup$
    – James E Hanson
    Commented Oct 19, 2023 at 2:09
  • $\begingroup$ @JamesHanson Naively I would mean the theory with primitives for elements and subsets of a real closed field, together with some axioms characterizing the sense in which the field is ‘real-closed’ based on second-order properties of the reals (like Dedekind completeness). $\endgroup$
    – Alec Rhea
    Commented Oct 19, 2023 at 2:12
  • $\begingroup$ So I'm not completely sure whether the integers are definable in that, but I do know that if you just take $\mathbb{R}$ as a field and add a new predicate for the reals, all Borel subsets of $\mathbb{R}$ becomes first-order definable, so you don't need too much to make a lot definable in $\mathbb{R}$. $\endgroup$
    – James E Hanson
    Commented Oct 19, 2023 at 2:19
  • $\begingroup$ @JamesHanson Interesting, thank you for the insight. $\endgroup$
    – Alec Rhea
    Commented Oct 19, 2023 at 2:22

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It sounds like there's some terminology confusion here. When $\mathcal{L}$ is a logic and $\mathcal{A}$ is a structure, "The $\mathcal{L}$-theory of $\mathcal{A}$" refers to the set $$\{\varphi\in\mathrm{Sent}_\mathcal{L}:\mathcal{A}\models_\mathcal{L}\varphi\}.$$

It sounds like you may mean something other than this, but it's not clear what you have in mind; for now I'll answer the question using this standard interpretation, but I'm happy to edit as the question is clarified.

Below I'll write "$Th_1(\mathcal{A})$" and "$Th_2(\mathcal{A})$" for the first- and second-order theories of a structure $\mathcal{A}$, respectively, and I'll let $\mathcal{R}$ denote the field of reals and $\mathcal{N}$ the semiring of natural numbers.

By definition, $Th_1(\mathcal{A})$ is always complete since for every sentence $\varphi$ we have either $\mathcal{A}\models\varphi$ or $\mathcal{A}\models\neg\varphi$. So it's not completeness that makes $Th_1(\mathcal{R})$ weak. We have to distinguish between $Th_1(\mathcal{R})$, which is prima facie complete but not obviously low-complexity, and the theory $\mathsf{RCF}$ which is prima facie low-complexity but is not obviously complete; of course, the theorem is that $Th_1(\mathcal{R})=\mathsf{RCF}$.

The same automatic completeness holds for $Th_2(\mathcal{A})$, or indeed $Th_\mathcal{L}(\mathcal{A})$ for any logic $\mathcal{L}$ which contains negation (if we omit negation a lot of things get much more complicated, see e.g. Shelah/Vaananen, Positive Logics).

OK, so how complicated is $Th_2(\mathcal{R})$? Well, as James said, it's extremely complicated even if we only use the "monadic" (= quantifying over only sets, as opposed to relations) version of SOL. In particular, the naturals are second-order definable in $\mathcal{R}$ since $r$ is a natural number iff for every set $X$, if $0\in X$ and $s\in X\implies s+1\in X$ then $r\in X$. So this makes $Th_2(\mathcal{R})$ at least as complicated as $Th_2(\mathcal{N})$, let alone $Th_1(\mathcal{N})$, which is itself massively more complicated than $\mathsf{PA}$.

However, even this is barely scratching the surface: $Th_2(\mathcal{R})$ tells us lots of things which even $Th_2(\mathcal{N})$ doesn't, such as whether or not the continuum hypothesis holds (it follows from sufficient large cardinals that $Th_2(\mathcal{N})$ can't be changed by forcing).

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    $\begingroup$ Another fascinating fact is that, if I recall correctly, the theory of $(\mathbb{R}, 2^{\mathbb{R}}, <, \in)$ computes true second-order arithmetic but doesn't interpret Peano arithmetic. $\endgroup$
    – James E Hanson
    Commented Oct 19, 2023 at 2:51
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    $\begingroup$ The second-order theory of $\mathbb R$ is in fact as complicated as third-order arithmetic: once you define $\mathbb N$ inside $\mathbb R$, you can also define binary expansion of reals which allows you to simulate subsets of $\mathbb N$ by elements of $\mathbb R$, and then subsets of $\mathbb R$ simulate families of subsets of $\mathbb N$. The other direction works as well. $\endgroup$
    – Emil Jeřábek
    Commented Oct 19, 2023 at 6:11
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    $\begingroup$ As a corollary to being able to simulate subsets of $\mathbb N$, you can talk about countable well-founded relations, because countable relations can be coded as sets of integers and second-order logic lets you identify well-foundedness. And similarly with simulating subsets of the powerset of $\mathbb N$. So you can construct $H_{\mathfrak{c}^+}$, the fragment of the set-theoretic universe consisting of sets hereditarily of cardinality $\le$ the continuum. This is enough to carry out a lot of set theory, e.g. identifying the existence of mice. $\endgroup$
    – Julia Williams
    Commented Oct 19, 2023 at 13:58

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