Does Euclidean geometry require a complete metric space?

Note: I'm not sure how to tag this question so please fix tags as appropriate and necessary.

This question relates to Tarski's axioms of Euclidean geometry, not Hilbert's axioms. It says on the same page (Theorem 3.1 -- see also this related question on Math.SE) that:

For every model of Euclidean Plane Geometry (using Tarski's axioms) there is a real closed field $R$ such that $M \cong R \times R$ as models.

From what I gather from nLab's page on real closed fields, these are strictly more general than the real numbers, for example the real algebraic numbers is also a real closed field.

However, the real algebraic numbers are not metrically complete. (I think.) In other words, not every Cauchy sequence converges.

Question: How can a real closed field, which is not metrically complete, serve as a model for Euclidean plane geometry?

In particular, how can the real algebraic numbers serve as a model for Euclidean geometry?

For example, such a field cannot express the ratio of the circumference of a circle to its diameter in Euclidean plane geometry since $\pi$ is transcendental and thus only contained in the metric completion of the rational numbers, but not the (real) algebraic completion.

One of Tarski's axioms is described on the nLab page as being a "Dedekind cut axiom expressed in first order terms". If I remember real analysis correctly, Dedekind cuts allow one to construct the metric completion of the rational numbers (the real numbers).

So if one of Tarski's axioms implies metric completeness (does it not? if not, why doesn't it?), then how can the real algebraic numbers, which aren't metrically complete (I think) serve as a model?

I'm not sure if the Cantor-Dedekind axiom is one of Tarski's axioms or the same as the aforementioned "Dedekind cut axiom". If Tarski's axioms don't require a metrically complete space, then is the Cantor-Dedekind axiom not generally valid for models of Tarski's axioms?

The answers to this related question seem to suggest that $\mathbb{Q}$ is insufficient for Euclidean geometry, which makes sense since $\mathbb{Q}$ is not real closed. However, it is unclear from the answers given if the problems with $\mathbb{Q}$ arise only because of the non-existence of algebraic irrational numbers (e.g. $\sqrt{2}$) or also because of the non-existence of transcendental irrational numbers, $\pi$. The latter would imply that the real algebraic numbers are also insufficient for Euclidean geometry.

Also, one of the axioms given for Euclidean plane geometry (on p. 171, M.2) in Agricola, Friedrich's Elementary Geometry is that the plane is a complete metric space. I had assumed that these axioms were just a (possibly less rigorous) rewording of Tarski's axioms, but if non-complete metric spaces also serve as models for Euclidean geometry, then it would seem that this axiom is an invention/addition of the authors, which perhaps should have been mentioned explicitly.

Solution 1:

Note: Since I'm not confident in this answer, because it comes from quoting a claim in a Wikipedia article which doesn't have a citation, and because I know next to nothing about mathematical logic, I am making it community wiki. This answer may be wrong, so please take it with a grain of salt.

The Wikipedia article about real closed fields (under "Model theory:...") says the following:

Euclidean geometry (without the ability to measure angles) is also a model of the real field axioms, and thus is also decidable.

That this is true seems to follow from Theorem 3.1 on the nLab page I mentioned earlier.

The key point in the above claim is the part in parentheses, "without the ability to measure angles". It seems that Tarski's axioms allow us to do everything we might want in Euclidean geometry except measure angles.

Not that this is impossible in all models of Tarski's axioms -- for instance it clearly is possible using $\mathbb{R}^2$ as a model. However, the fact that $\pi$ is transcendental and not algebraic seems to imply that it isn't possible to measure (all) angles using the real algebraic numbers as a model for Tarski's axioms of Euclidean geometry.

(This is because the definition of angles in terms of radians uses $\pi$.)

Thus, I wold conclude from this that the answer to my question likely is:

Metric completeness is required in a model for Euclidean geometry if and only if one needs to measure angles. If one does not need to measure angles, then it is unnecessary.

Note: It occurred to me recently that my question is similar in spirit to one asked earlier by someone else: In algebraic geometry, why do we use $\mathbb C$ instead of the algebraic closure of $\mathbb Q$?

In particular, the answers to that question seem to indicate that, besides making measuring angles possible, metric completeness can sometimes make geometry more convenient.

Also, I have to imagine that Tarski proved both the Lefschetz principle and his axioms for Euclidean geometry is not a coincidence. Namely both involve statements about the equivalence of (real) algebraically closed fields in first order logic.