Mancosu - Philosophy of Mathematical Practice (Oxford, 2008).pdf

purity of method in hilbert’s grundlagen 223and these planes intersect in a straight line. Cast in Hilbert’s system, it is easilyproved using just the full (i.e. planar and spatial) incidence and order axioms(Groups I and II). However, the theorem has a restricted version, wherethe triangles in question both lie in the same plane.²⁶ This version is not soeasily visualizable, neither is it especially easy to prove. More importantly, thestandard proof goes via the unrestricted version, using a point outside the planeof the triangles to reconstruct a three-dimensional Desargues’s arrangementwhereby the three-dimensional version of the theorem can be applied. Inshort, this proof, too, calls on all the axioms of I and II, even though thetheorem itself appears to involve only planar concepts, i.e. is concerned onlywith the intersection of lines in the same plane. An indication of the standardway of proving the planar version of Desargues’s Theorem is given by Fig. 8.1.Intuition might seem to play a role here, too, since it relies on projection ofthe original triangles up from the plane they lie in; in short, one can ‘see’that planar Desarguesian configurations are reflected in spatial Desarguesianconfigurations and conversely.This situation is remarked upon by Hilbert in his 1898/1899 lectures asraising a purity problem. Hilbert writes:I have said that the content of Desargues’s Theorem is important. For now howeverwhat’s important is its proof, since we want to connect to this a very importantconsideration, orratherline of enquiry. The theorem is one of plane geometry; theproof nevertheless makes use of space. The question arises whether there is aproof which uses just the linear and planar axioms, thus I 1–2, II1–5. Thus hereSB′ OB OB′C′A′BCAOFig. 8.1. Diagram for the usual proof of the Planar Desargues’s Theorem, taken from:Hilbert and Cohn-Vossen (1932), p.108, Fig. 134. ABC and A ′ B ′ C ′ are the twotriangles in the same plane, and S is an auxilliary point chosen outside this plane.²⁶ Both versions have fully equivalent converses, where it is assumed that the three points ofintersection lie on a line, and it is then shown that the lines AA ′ , BB ′ , CC ′ meet in a point.