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While reading a paper1 about the history of modern logic, I learned that some opinions (about deductive/axiomatic mathematics) typically attributed to David Hilbert can be traced back to Moritz Pasch. After googling for Moritz Pasch, I was surprised to learn that he had found important implicit assumptions in Euclid missing from the axioms/postulates. I read on Wikipedia that both Pasch's theorem and Pasch's axiom cannot be derived from Euclid's postulates.

Are there simple models similar to elliptic and hyperbolic geometry for the parallel postulate that allow to illustrate this fact in a simple way?

1Jose Ferreiros: The Road to Modern Logic - An Interpretation, The Bulletin of Symbolic Logic, Vol. 7, No. 4 (Dec., 2001), pp. 441-484. Available here, JSTOR, DOI: 10.2307/2687794.

The Amplitwist
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Thomas Klimpel
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    It seems to me that whether Pasch's theorem is a theorem of plane geometry or not depends on what one considers plane geometry to be; the theorem depends critically on a notion of order which isn't even part of geometry as Euclid defines it. AFAIK the only less than/greater than comparison in Euclid's axioms is in the parallel postulate, where the angles 'less than right angles' define the side on which non-parallel lines meet, and it's not even clear that that can be used to define a linear notion of less than, greater than, or between. – Steven Stadnicki Apr 05 '13 at 22:12
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    As such, violating Pasch's theorem seems as simple as defining a new 'between' relation of one's own and saying by fiat that "B between (A, C)" and "C between (B,D)" but not "B between (A,D)". To invalidate this sort of trivial counterexample then a more explicit definition of the linear-order relationship would be needed, and it seems plausible that any useful definition that correlates with an inherent notion of betweenness would then allow the derivation of Pasch's theorem. – Steven Stadnicki Apr 05 '13 at 22:15
  • @StevenStadnicki OK, I see. I guess that's the reason why "Pasch's theorem" cannot be derived, but is less important than "Pasch's axiom". – Thomas Klimpel Apr 05 '13 at 22:28

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There are no simple models. To violate Pasch's Axiom in a Hilbert-type setup, we need to use a discontinuous solution of the Cauchy functional equation $f(x+y)=f(x)+f(y)$. Such a discontinuous solution requires (part of) the Axiom of Choice.

In ZF with added axiom that every set of reals is Lebesgue measurable, the Pasch Axiom is a theorem of a Hilbert-style axiomatization that leaves out the Pasch Axiom.

Remark: The first construction of a non-Paschian geometry that otherwise satisfies the full set of Hilbert's axioms is due to Szmielew. A proof that such a geometry must be of the Szmielew type was given by Adler.

Lee Mosher
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André Nicolas
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    What happens if you also throw out Dedekind completeness so as to introduce gaps or holes into the lines? Is it then possible to give an explicit model in which Pasch fails (note that Euclid also does not mention anything like Dedekind completeness, so a model which jettisons this axiom may still be considered to be "model of Euclid's postulates" that is still a "plane geometry")? – The_Sympathizer Oct 26 '14 at 06:40
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    That raises an interesting question, which could be fruitful, about what we can say over Euclidean fields. Maybe the answer is simple, but I know the (older) literature pretty well, and do not recall anything. I tempted to start exploring! – André Nicolas Oct 26 '14 at 15:30
  • Regarding continuity (eliminating gaps in lines), Euclid's geometry only requires that we work over an (fittingly named) Euclidean field, which is an ordered field consisting of all the rationals and the square roots of all its positive elements. That is because Euclid only allows figures constructed with straightedge and compass, and constructions with these tools are equivalent to solving algebraic equations of degrees 1 or 2. – Claudio Buffara Jan 19 '21 at 12:52