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).

  • $\begingroup$ Is it possible that you are confused between "complete" as in "completeness theorem" as a property of logic, and "complete" as in "Dedekind-complete" or "Cauchy-complete" as properties of fields? $\endgroup$ Commented Jan 13, 2020 at 21:58
  • $\begingroup$ No, I mean in the logical sense, that Complete Ordered Fields are unique up to isomorphism is stated for instance, here: math.stackexchange.com/questions/269353/… (citing Spivak's Calculus) $\endgroup$ Commented Jan 13, 2020 at 22:03
  • $\begingroup$ Or here where it is proved that "Let Fb be a complete ordered field. Then there exists an isomorphism of Fb onto the Dedekind field F": spot.colorado.edu/~baggett/appendix.pdf $\endgroup$ Commented Jan 13, 2020 at 22:06
  • $\begingroup$ My confusion is how something could be unique up to isomorphism but (maybe) incomplete. $\endgroup$ Commented Jan 13, 2020 at 22:08

1 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 $(\mathbb{R};+,\times)$.

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!

  • $\begingroup$ I think what OP has in mind here is a second-order theory, not RCF. $\endgroup$ Commented Jan 13, 2020 at 22:11
  • $\begingroup$ @EricWofsey Given that they talk about incompleteness, I got the opposite impression - that they're asking why the incompleteness of first-order PA doesn't extent to $Th(\mathbb{R})=RCF$. $\endgroup$ Commented Jan 13, 2020 at 22:12
  • $\begingroup$ The title mentions "complete ordered fields" and the reals being unique up to isomorphism, though. $\endgroup$ Commented Jan 13, 2020 at 22:13
  • $\begingroup$ @EricWofsey And the second half of the title is "anything that can prove Peano Arithmetic is incomplete?," which clearly doesn't apply to second-order Peano arithmetic. We'll see whether this satisfies the OP. $\endgroup$ Commented Jan 13, 2020 at 22:15
  • $\begingroup$ (That is, I think the OP is mentioning the uniqueness of complete ordered fields to emphasize that $Th(\mathbb{R})$ is a fortiori complete, and confused why that doesn't contradict the complexity of the theory of one of its substructures. I've rephrased my answer to hopefully be clearer.) $\endgroup$ Commented Jan 13, 2020 at 22:23

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