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By a Tarski space I understand a mathematical structure $(X,B,\equiv)$ consisting of set $X$, a betweenness relation $B\subseteq X^3$ and a congruence relation ${\equiv}\subseteq X^2\times X^2$ satisfying all Tarski's axioms except for the Euclid Axiom, the Axiom of Continuity, and the Dimension Axioms. In other words, Tarski spaces should satisfy seven Tarski's axioms: (1) Reflexivity of Congruence, (2) Identity of Congruence, (3) Transitivity Congruence, (4) Identity of Betweenness, (5) Five Segments, (6) Segment Construction, (7) Inner Pasch Axiom.

As shown in the book of Schwabhäuser, Szmielew and Tarski, a substantial portion of familiar (absolute) geometry holds in Tarski spaces of dimension $\ge 2$. However the proofs of some 2-dimensional results, for example, the three circles and common chords, require one additional dimension. Another example is constructing an equilateral triangle, which can be done in 3-dimensional Tarski space but is problematic in dimension 2. This motivates my

Question. Does every Tarski plane embed into a 3-dimensional Tarski space?

Or the situation with the three circles and common chords resembles the situation with the Desargues Theorem, which is true for higher dimensional projective spaces, but in dimension 2 has counterexamples, called non-Desarguesian planes?

Added in Edit: I have just realized that Tarski planes are exactly the Hilbert planes, whose theory is well described in two excellent textbooks of Greenberg and Hartshorne. Hilbert planes were classified by Pejas in 1961. For Hilbert planes satisfying the axiom (E), the Pejas classification is presented in Theorem 43.7 of Hartshorne's textbook. The axiom (E) is the Circle-Circle Intersection Property. The Pejas classification implies that Tarski (=Hilbert) planes with the axiom (E) indeed embed into higher-dimensional Tarski spaces. As written on page 423 of Hartshorne's textbook, Pejas' classification implies that the axiom (E) is equivalent to the Line-Circle intersection property (LCI). Therefore, Tarski (=Hilbert) planes with (LCI) also embed into 3-dimensional Tarski spaces. So, it remains to understand whether the Pejas classification of Hilbert planes implies that every Hilbert plane embeds into a 3-dimensional Tarski space. But I cannot even find a good formulation of the full Pejas characterization in the literature (the original Pejas paper in Mathematische Anallen is not accessible, at least for me).

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  • $\begingroup$ Do you know anything about the completions of this theory in various dimensions? $\endgroup$
    – James E Hanson
    Commented Jan 14, 2023 at 17:27
  • $\begingroup$ Depending which kind of completion you have in mind. Every Tarski space has a natural uniformity and we can take the completion by this uniformity. If the Tarski space satisfies the Parallel Axiom and Archimedean Axiom, then the completion carries the structure of a real Hilbert space. But without those axioms I do not know what will happen with the completion and whether it will have a structure of a Tarski space. $\endgroup$
    – Taras Banakh
    Commented Jan 14, 2023 at 18:42
  • $\begingroup$ Sorry, what I mean is completions of the first-order theory, not completions of the spaces themselves. $\endgroup$
    – James E Hanson
    Commented Jan 14, 2023 at 19:24
  • $\begingroup$ @JamesHanson Ok, then if the completion means adding some axioms that make the theory complete, then yes, the all the Tarski (first-order) axioms seem to produce a complete theory (of finite powers of real closed ordered fields). $\endgroup$
    – Taras Banakh
    Commented Jan 14, 2023 at 20:43
  • $\begingroup$ Doesn't that give you a way to embed a Tarski plane into a 3-dimensional Tarski space by taking the third power of the corresponding real closed ordered field? $\endgroup$
    – James E Hanson
    Commented Jan 14, 2023 at 21:15

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