In David Hilbert's 1899 Grundlagen der Geometrie, Hilbert gives a rigorous axiomatization of Euclidean geometry. As I understand it, some of Hilbert's axioms must be expressed in second order logic (for example his "Euclid's axiom") and as his system contains some notion of arithmetic (this is perhaps not correct?), by Gödel's first incompleteness theorem this means that Hilbert's axiomatic system is incomplete (meaning that there exist statements expressed by the vocabulary of the system that cannot be proven to be true) and by Gödel's second incompleteness theorem, his system cannot demonstrate its own consistency.
Later, Tarski gave his own axiomatization of Euclidean geometry that is entirely in first order logic so by Gödel's completeness theorem, it can demonstrate its consistency, it is decidable and complete.
Is the above accurate and if so, is Tarski's axiomatic system still inferior in some ways to Hilbert's (with a lack of expressive power or doesn't cover the entire Euclidean geometry as we understand it)?
Best Answer
Hilbert's axioms predate the development (or at least the wide adoption) of fully symbolic logic, so they are expressed in partially informal language -- though Hilbert strove to make them as precise as he could.
They include the Axiom of Archimedes, formulated in language that presupposes that the natural numbers are already known. As such, if we want to formalize the system in a modern sense, we'd need to add in some machinery for talking about integers, and it would be somewhat hard to avoid making that machinery powerful enough that Gödel's theorem would apply to it.
Furthermore, there's also an Axiom of Completeness which attempts to claim directly that the structure we're speaking about is maximal in a certain technical sense. In the modern view of formal theories, this hardly qualifies as an "axiom" at all, because it doesn't assert the truth of any formula interpreted inside the language of the theory itself. In modern terms, it appears to try to say that every existential formula of a certain shape that is true in some model of the other axioms must itself be elevated to axiomatic status.
However, it is not clear that "formulas that are true in some model of the other axioms" are even recursively enumerable -- which means that Hilbert's axioms, interpreted in this way, are not effective. In other words there's no systematic way to check whether a purported proof is valid or not. Therefore Gödel's incompleteness theorem does not apply to Hilbert's axioms. It seems at least plausible that if we interpret them inside set theory in the above sense, they do have $\mathbb R^3$ as their only model up to isomorphism. (That is, whatever the set theory in question considers $\mathbb R^3$ to be).
Tarski's axioms for geometry were created after formal logic was better developed. They form a genuine, effectively axiomatized, first-order theory, with no ad-hoc appeals to integers, sets, or model theory. They manage to be a complete theory because they are not strong enough to express or simulate arithmetic.
The price paid for the completeness in Tarski's case is that the language the axioms are formulated in is not expressive enough to speak of even finite sets of points, or for example general polygons. Every theorem has to be proved separately for triangles, quadrilaterals, pentagons, and so forth.
This restriction is unavoidable for a complete theory, because as soon as we extend the language with a way to speak of finite sets of points and lines in a reasonable way, we can use those sets as proxies for natural numbers (each set representing the number of points in it) and begin to speak about enough arithmetic that Gödel's incompleteness theorem will apply to it.
As mentioned in the comments, this is a complete misunderstanding of what the completeness theorem says. Tarski's geometry cannot even speak about its own consistency, much less prove it.