FOM: Re: Axiom of infinity
shipman at savera.com
Thu Jul 20 14:57:56 EDT 2000
Oops, I got that backwards -- every Pappian plane is Desarguesian and every
finite Desarguesian plane is Pappian so you have to take the axioms for a
projective plane with Desargues's axiom and the negation of Pappus's axiom.
The way I said it below is inconsistent.
For the record:
Projective plane axioms
1) For any two distinct points there is a unique line incident with both
2) For any two distinct lines there is a unique point incident with both
3) There exist four distinct points no three of which are on any line
Definition: a triangle is an ordered triple of points which are not all on a
line. In the following, all labeled points are assumed distinct.
If ABC, A'B'C' are two triangles such that AA', BB', CC' meet in a point, the
intersections of BC with B'C', of CA with C'A', and of AB with A'B' are all in
If A, B, C are in one line and A', B', C' in another line, the intersections
BC' with B'C, of CA' with C'A, and of AB' with A'B are all in one line
Joe Shipman wrote:
> Another example, equivalent to the one below, would be the conjunction of
> the axioms for a projective plane with Pappus's axiom and the negation of
> Desargues's axiom.
> -- JS
> Joe Shipman wrote:
> > Simpson asks for a simple, short axiom of infinity that doesn't
> > interpret one of the three mentioned by Baldwin (discrete order, dense
> > order, pairing function).
> > How about the conjunction of the axioms for a division ring augmented by
> > the negation of the commutativity axiom for multiplication? By
> > Wedderburn's theorem, a finite division ring is a field, so the only
> > models of this sentence would be infinite. This is a hard theorem to
> > prove, so there can't be an easy proof that the sentence interprets one
> > of the three infinity axioms Baldwin names.
> > -- Joe Shipman
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