An essay on the foundations of geometry · Bertrand Russell
FOOTNOTES:
Let _O_, _P_, _Q_ be the three points whose projective relation is required. Then we have given us the three straight lines _PQ_, _QO_, _OP_. Metrically, the relation between these points is made up of the area, and the magnitude of the sides and angles, of the triangle _OPQ_, just as the relation between two points is distance. But projectively, the figure is unchanged when _P_ and _Q_ travel along _OP_ and _OQ_, or when _OP_ and _OQ_ turn about _O_ in such a way as still to meet _PQ_. This is a result of the general principle of projective equivalence enunciated above (§§ 108, 109). Hence the projective relation between _O_, _P_, _Q_ is the same as that between _O_, _p_, _q_ or _O_, _P′_, _Q′_; that is, _p_, _q_ and _P′_, _Q′_ lie in the plane _OPQ_. In this way, any number of points on the plane may be obtained, and by repeating the construction with fresh triads, every point of the plane can be reached. We have to prove that, when the plane is so constructed, the straight line joining any two points of the plane lies wholly in the plane. It is evident, from the manner of construction, that any point of _PQ_, _OP_, _OQ_, _OP′_ or _OQ′_ lies in the plane. If we can prove that any point of _pq_ lies in the plane, we shall have proved all that is required, since _pq_ may be transformed, by successive repetitions of the same construction, into any straight line joining two points of the plane. But we have seen that the same plane is determined by _O_, _p_, _q_ and by _O_, _P_, _Q_. The straight lines _PQ_, _pq_ have, therefore, the same relation to the plane. But _PQ_ lies wholly in the plane; therefore _pq_ also lies wholly in the plane. Hence our axiom is proved.
Text from Project Gutenberg, public domain in the United States.