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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.

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    $\begingroup$ I don't understand what you find puzzling about there being a strange or unexpected model of a first-order axiom system. Isn't asking "how can there by this weird model of axiomatized Euclidean plane geometry?" like asking "how can there be nonstandard (e.g. uncountable) models of first-order Peano arithmetic?" The answer is: because the axioms allow it, and it's the business of model theory to discover such allowances. The error is in thinking the axioms entailed metric completeness: the model theory shows they don't. $\endgroup$ – symplectomorphic Mar 8 '17 at 19:17
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    $\begingroup$ The point is that in Tarski's axioms, any point you can construct has coordinates which are real algebraic over any points used to construct it. This is why real closed fields are the natural level of generality. Tarski's axioms don't directly talk about things like the circumferences of circles. $\endgroup$ – Qiaochu Yuan Mar 8 '17 at 19:20
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    $\begingroup$ PS: You're engaging in equivocation when you write "if non-complete metric spaces also serve as models for Euclidean geometry, then it would seem ..." Those spaces serve as models for Tarki's axiomatization of Euclidean geometry, not Euclidean geometry, period. It seems to me bizarre to assume that anytime, in any context, an author says "Euclidean geometry," without further qualification, he or she must be referring to models of Tarksi's axioms. The author could be giving a rival axiomatization. Or the author could be referring to the standard model of $\mathbb{R}^2$. Etc. $\endgroup$ – symplectomorphic Mar 8 '17 at 19:32
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    $\begingroup$ William: I don't understand all the flack you are getting. I think this is a great question. It is well researched, shows an attempt to reconcile different mathematical intuitions, and demonstrates courage to branch out of your area of expertise. Furthermore, it isn't written in a tone which argues that there is something wrong with the mathematics. I wish more questions were like this. $\endgroup$ – Kyle Mar 9 '17 at 0:31
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    $\begingroup$ @William: What is your background in model theory? Do you know the compactness theorem? $\endgroup$ – Kyle Mar 9 '17 at 0:35
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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.

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