Can an uncountable model of Peano Arithmetic be recursive?

Question

Can an uncountable model of Peano Arithmetic be recursive?

What does it mean for an uncountable model to be recursive? Well, we represent the elements of the model using real numbers instead of natural numbers, and assume $+, \times, ^{-1}$ are computable functions and $\ge$ is a computable relation on the real numbers. (In particular, we could assume we are using lambda calculus, and add symbols for $+, \times, \ge, ^{-1}$, as well as a symbol for each computable real number. Or you could use some other model of Real computation.)

So an uncountable model $M$ of Peano Arithmetic is recursive if $\mathbb N_M \subseteq \mathbb R$ is a computable set, and $+_M, \times_M$ are computable.

Answer 1

I will not be able to give a full answer, but rather attempt to make the question precise and give some partial answers.

First, I shall side with Andrej Bauer's point in the comments that there ought to be a consensus what computability ought to mean here. Bringing in e.g. the BSS-model would essentially amount to a model-theoretic investigation (is there an interpretation of this theory in that theory) rather than a question of computability.

We are thus asking about represented spaces (in the sense of computable analysis) and computable functions. More precisely, we ask:

Question: What represented spaces $\mathbf{N}$ with computable operations $\times, + : \mathbf{N} \times \mathbf{N} \to \mathbf{N}$ and computable constants $0, 1 \in \mathbf{N}$ are there such that $(\mathbf{N},0,1,+,\times)$ is a model of Peano arithmetic? We shall call these "Type-2 computable models".

First Observation: Each countable ultrapower of $\mathbb{N}$ yields a Type-2 computable model (which is uncountable).

This is just because ultrapowers trivially preserve computability of functions, they only mess with relation symbols. Of course the equality of an ultrapower is about as complicated as the ultrafilter we are using. So this does not even give us Borel equality.

Second Observation: There are Type-2 computable non-standard models with decidable equality (which are countable).

Pick your favourite countable PA-model $M$, and let $p \in 2^\omega$ denote its atomatic diagram. Now define a representation $\rho_M$ of $M$ as follows:

$01^n0^\omega$ is a name for the standard $n \in M$. $11^n0p$ is a name for the $n$-th non-standard number in $M$ according to the injective enumeration we picked for defining $p$.

Equality is trivially decidable. For $+$ and $\times$, either both inputs are standard numbers, and thus the output is standard again; or we can figure out what the output is with the help of $p$ which a non-standard input provides. QED

As a side remark: This shows that Tennenbaum's theorem is relying on the tacit assumption that all numbers are computable.

Now having decidable equality implies being countable, so if we want uncountable models, we do have to relax our demands a bit. The natural requirement would seem to be being Hausdorff, i.e. having equality being co-ce. So the actual (and still open) question would seem to be:

Question: Are there uncountable Hausdorff Type-2 computable models of PA?