# A model of geometry with the negation of Pasch’s axiom? [duplicate]

How would a geometry with all the usual axioms of Euclidean geometry, except that instead of Pasch’s axiom, we take it negation, look like?

## marked as duplicate by colormegone, hardmath, Claude Leibovici, user91500, user228113 Jul 10 '16 at 10:30

• What do you take as "the usual axioms of Euclidean geometry"? Hilbert's? – Blue Feb 5 '14 at 23:05

For elementary geometry without the Pasch Axiom, every model is isomorphic to a Cartesian plane over a formally real Pythagorean semi-ordered field $\mathcal{F}$.

More can be said if our geometry has the full second-order continuity axiom. For then $\mathcal{F}$ is (as a field) isomorphic to the reals.

Now let $f:\mathbb{R}\to \mathbb{R}$ be a non-linear solution of the Cauchy functional equation $f(x+y)=f(x)+f(y)$, where $f$ is onto and $0\lt f(1)$. Define the order relation $\lt^\ast$ by $x\lt^\ast y$ if $f(x)\lt f(y)$. Then under $\lt^\ast$ and ordinary addition, the reals form an ordered group. The Cartesian plane over the reals, with order relation given by $\lt^\ast$, is a non-Paschian model of geometry with second-order continuity.

References: The use of the Cauchy functional equation to produce a non-Paschian geometry is due to Szczerba, Independence of Pasch's Axiom (1970). A proof that all non-Paschian geometries with full continuity axiom are of the shape described by Szczerba was given by Adler, Determinateness and the Pasch Axiom, 1971.

Remark: The second result has an interesting consequence. Under suitable conditions, we can have a model of ZF in which all sets of reals are Lebesgue measurable. In such a model, any model of the Hilbert second-order geometry minus the Pasch Axiom is automatically Paschian!