Is the theory of the complex field along with the real numbers $2^{\aleph_0}$-categorical?

Question

I know that the theory of the complex field $(\mathbb{C};+,-,*,0,1)$ is $2^{\aleph_0}$-categorical, while the theory of the real field $(\mathbb{R};+,-,*,0,1)$ is not. However, consider the structure $(\mathbb{C};+,-,*,0,1,R)$, where $R$ is a unary predicate that picks out the real numbers. Is the complete theory of that structure $2^{\aleph_0}$-categorical?

Answer 1

The first thing everyone should know about categoricity for first-order theories is that it is an extremely rare phenomenon, and for all infinite cardinals $\kappa$ (greater than or equal to the cardinality of the language), we have clear and restrictive criteria for when a theory $T$ is $\kappa$-categorical. Let's assume the language is countable and $T$ is complete. Then:

• The Ryll-Nardzewski theorem says (in one formulation) that $T$ is $\aleph_0$-categorical if and only if there are only finitely many complete types over any finite set in a model of $T$.
• The Baldwin-Lachlan proof of Morley's categoricity theorem gives that $T$ is $\kappa$-categorical for some uncountable $\kappa$ if and only if $T$ is $\kappa$-categorical for every uncountable $\kappa$ if and only if $T$ is $\aleph_0$-stable with no Vaughtian pairs. We say $T$ is uncountably categorical in this case.

The condition of $\aleph_0$-stability means that there are only countably many complete types over any countable set in a model of $T$. It implies that $T$ is a stable theory. In particular, no theory which defines a order on some infinite subset of a model can be uncountably categorical.

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Let $T$ be the complete theory of $(\mathbb{C};+,*,0,1,R)$, where $R$ picks out the real numbers. This theory defines the linear order on $R$, by $x\leq y$ iff $\exists z\, (R(z)\land z^2 = y-x)$. As Mark Kamsma notes in the comments, this immediately implies that $T$ is unstable, so it is not uncountably categorical.

We can also see that there are uncountably many complete types over $\mathbb{Q}$: for each irrational real number $r$, the complete type $\mathrm{tp}(r/\mathbb{Q})$ includes the information of which cut in $\mathbb{Q}$ contains $r$, so distinct real numbers have distinct complete types over $\mathbb{Q}$. Thus $T$ is not $\aleph_0$-stable and not uncountably categorical.

In fact, $T$ is as far from being $2^{\aleph_0}$-categorical as possible: it has the maximal number of non-isomorphic models of cardinality $2^{\aleph_0}$: $2^{2^{\aleph_0}}$. This actually follows from general (un)stability theory - it's a theorem of Shelah that for every uncountable cardinal $\kappa$, every countable unstable theory has $2^{\kappa}$-many models of cardinality $\kappa$ up to isomorphism.

But to see this a bit more concretely, let's note that a model of $T$ is an algebraically closed field $C$ in which $R$ names a real closed subfield such that $R[i] = C$. In this recent answer, I sketched how to construct $2^{2^{\aleph_0}}$-many real closed fields of cardinality $2^{\aleph_0}$ up to isomorphism. If $\mathcal{R}$ is one of these real-closed fields, then $(\mathcal{R}[i];+,*,0,1,\mathcal{R})$ is a model of $T$, and any isomorphism between two such models of $T$ induces an isomorphism between their distinguished real closed subfields. So this construction produces $2^{2^{\aleph_0}}$-many non-isomorphic models of $T$.