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This is really a question about references. The entry in Russian Wikipedia about Hilbert's axioms states, in particular, that completeness of Hilbert's system was proven by Tarski in 1951. The reference is to the Russian encyclopedia of elementary mathematics, to which I don't have access, and I somehow am not able to find any references to this statement in the literature.

I have two questions:

  1. The continuity axioms are not first-order-logic statements. What does then completeness mean in this situation?

  2. Does anybody know a reference to the original proof by Tarski or any other proof of this statement, for that matter?

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  • $\begingroup$ For 1, I'd say that Hilbert's Foundations of Geometry was a key step in the formalist vision of math, but his notion of completeness was vague, given that he was writing before the formalization of set theory and the standardization of first-order logic. $\endgroup$
    – user44143
    Commented Feb 4, 2020 at 19:28
  • $\begingroup$ You should not trust Russian Wikipedia. It is written by uneducated people who accidentally seized power. Almost everything written there on mathematics contains a substantial amount of absurdity. Here is an example of a debate on what was written about the foundations of mathematics: ru.wikipedia.org/wiki/… $\endgroup$ Commented Feb 4, 2020 at 20:25
  • $\begingroup$ @Matt F. The question is though about the statement and the proof by Tarski, not Hilbert. $\endgroup$
    – R. Matveev
    Commented Feb 5, 2020 at 7:56
  • $\begingroup$ @Sergei Akbarov. I did not express any opinion about the content of any Wikipedia page. $\endgroup$
    – R. Matveev
    Commented Feb 5, 2020 at 7:57
  • $\begingroup$ The question currently mentions “the completeness of Hilbert’s system” and “not first-order” statements, as if asking about a statement by Hilbert and proof by Tarski. If you’re only interested in the Tarski part, it would help to clarify that. $\endgroup$
    – user44143
    Commented Feb 5, 2020 at 12:44

1 Answer 1

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The original is Alfred Tarski's book "The completeness of elementary algebra and geometry", which was due to appear in 1940 but never made it into print because of the outbreak of WW2. An edition appeared after all in 1967 (Institut Blaise Pascal, Paris), but is not easy to come by.

Essentially the same argument is presented in Tarski's 1948/51 "A decision method for elementary algebra and geometry", available here.

Tarski uses an axiomatic setup of elementary geometry different (but equivalent) to Hilbert's, using only points (not lines or planes) as primitive terms, and two relation symbols, $B$ and $D$ (ternary and quaternary, respectively). $Bxyz$ signifies that the point $y$ is on the "line" $xz$ between the points $x$ and $z$ ("betweenness"), while $Dxyzw$ means that the distance between $x$ and $y$ equals that between $z$ and $w$ (the "equidistance" relation). Here, $x$, $y$, $z$ (and $w$) are allowed to coincide.

Full continuity is the second order axiom $\forall_{X}\forall_{Y}((\exists_{a}\forall_{x\in X}\forall_{y\in Y}Baxy)\rightarrow(\exists_{b}\forall_{x\in X}\forall_{y\in Y}Bxby))$, where $X$ and $Y$ are variables ranging over sets of points.

First order continuity is weaker, and is expressed as an axiom schema where $X$ and $Y$ are given as $\lbrace x\mid \phi(x)\rbrace$ and $\lbrace y\mid \psi(y)\rbrace$, respectively, for arbitrary first order formulas $\phi(x)=\phi(x,p_{1},\cdots,p_{n})$ and $\psi(y)=\psi(y,p_{1},\cdots,p_{n})$ that are allowed to contain parameters $p_{1},\cdots,p_{n}$.

A special case of first order continuity is the Circle Axiom, by which a line that contains an interior point of a circle (in the same plane) must meet that circle.

Completeness is the statement that any model $\mathfrak A$ of the axioms of elementary $n$-dimensional geometry without continuity is isomorphic to $K^{n}$ (with the obvious interpretations for the $B$ and $D$ relations) for a Pythagorean ordered field $K$ (that is $K\models\forall_{a}\forall_{b}\exists_{c}(a^{2}+b^{2}=c^{2})$), uniquely determined by $\mathfrak A$ up to isomorphism.

Under full continuity, $K$ must be $\mathbb{R}$, under first order continuity $K$ must be real closed, and for the Circle Axiom $K$ merely needs to be Euclidean (i.e., $K\models\forall_{a}\exists_{b}(a=b^{2}\vee -a=b^{2})$).

This is the content of the Representation Theorem, Th. I, (16.15) in "Metamathematische Methoden in der Geometrie" by W. Schwabhäuser, W. Szmielew and A. Tarski, Springer Hochschultext, 1983, an excellent reference for the metamathematics of elementary geometry (in German).

Edit: Let me add a few comments.

  1. The statement above that Tarki's setup is equivalent to Hilbert's is rather imprecise, as noted by Matt F. and others. Tarski works in first order logic, while a formalization of Hilbert's system is at least unclear. (Still, the axioms in Hilbert's axiom groups I-IV can be derived from Tarski's axioms, as shown in the Schwabhäuser, Szmielew, Tarski text).
  2. For the same reason, it is not clear what it would mean for Hilbert's system to be complete (in the modern sense), and I do not claim that "completeness" of Hilbert's system follows from that of Tarski. Hilbert includes a "completeness axiom", to the effect that his "model" of the axioms in groups I-V (where V is archimedeanity) cannot be extended to a "model" with a larger universe.
  3. To add to the confusion, my use of the word Completeness above (in the body of the answer, in reference to the Representation Theorem) was also unfortunate. Tarski has shown that the first order theory of real closed fields is complete (in the modern sense), and that, as a result, the same goes for the theory of $n$-dimensional elementary geometry (based on Tarski's axioms, with the first order continuity axiom schema included).
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    $\begingroup$ Why do you say that Tarski's setup is equivalent to Hilbert's? Tarski's is first-order and formalized, Hilbert's is not first-order and by 21st-century standards rather informal. I wouldn't say "different (but equivalent) to Hilbert's", when "based on Hilbert's ideas" seems more accurate. $\endgroup$
    – user44143
    Commented Feb 4, 2020 at 19:22
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    $\begingroup$ @Matt F. Hilbert's axioms, as set forth in his Grundlagen, are, although informal, no less precise. Formalizations of his principles can be deduced in Tarski's system. $\endgroup$ Commented Feb 4, 2020 at 20:05
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    $\begingroup$ @MatthévanderLee How is your comment responsive to Matt's point that Hilbert's system is not first-order? The archimedean and completeness axioms in Hilbert's system fall outside Tarski's system (and so cannot be deduced therein). $\endgroup$ Commented Feb 4, 2020 at 20:40
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    $\begingroup$ @SergeiAkbarov. Any extension of ZFC or any other set theory capable of modelling $\mathbb{N}$ will be incomplete by Gödel's Theorem, while Tarski's system, which is based on Hilbert's as explained above, is complete. $\endgroup$
    – R. Matveev
    Commented Feb 5, 2020 at 8:29
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    $\begingroup$ I don't want to continue this discussion in chat. What you explain here, is clear without your explanations. What is not clear (and sounds strange) are declarations like this: $$ \text{> Hilbert system is equivalent to Tarski, as explained above.} $$ You should give a definition of this equivalence before writing this, @R.Matveev. And giving advices like this: $$ \text{> For more details, see the answer and the discussion.} $$ $\endgroup$ Commented Feb 5, 2020 at 13:38

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