FOM: Re: Axiom of infinity
Joe Shipman
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.
Desargues axiom
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
one line
Pappus axiom
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
-- JS
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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