How can it be true both that Complete Ordered Fields are unique up to isomorphism and that anything that can prove Peano Arithmetic is incomplete?

Question

The real numbers include the natural numbers which presumably satisfy the Peano axioms, I don't know how you could be a strong enough theory to prove the existence of a set that satisfies the Peano axioms but have the axioms be not strong enough to prove its own incompleteness (and presumably if it's not complete, it is not unique up to isomorphism).

Answer 1

The theory of $(\mathbb{R};+,\times)$ is indeed consistent, complete and computably axiomatizable - it happens to be exactly the theory of real closed fields - and of course $\mathbb{N}\subseteq\mathbb{R}$. However, $\mathbb{N}$ is not a definable subset of $(\mathbb{R}; +,\times)$! This prevents the logical complexity of $(\mathbb{N};+,\times)$ from being inherited by $(\mathbb{R};+,\times)$: the latter is bigger, but not more complicated.

What is true - and is what you're gesturing at when you write "anything that can prove Peano Arithmetic is incomplete" - is that if $T$ is any computably axiomatizable theory such that some model of $T$ has a definable copy of $(\mathbb{N}; +,\times)$ then $T$ is not complete, and similarly any consistent computably axiomatizable theory which interprets the (very weak) theory of Robinson arithmetic is not complete. But none of that applies here since we don't have definability of $\mathbb{N}$ in $(\mathbb{R};+,\times)$.

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It might be easier to consider a more algebraic example: $(\mathbb{C}; +,\times)$ is algebraically much simpler than $(\mathbb{R};+,\times)$ despite being a larger field. For example, the set of polynomials (in any number of variables) which have a zero is much simpler to describe over $\mathbb{C}$ than over $\mathbb{R}$. Similarly, while $\mathbb{R}$ has no nontrivial automorphisms at all there are lots of automorphisms of $\mathbb{C}$ - including ones which move reals!