WILLIAMS. — GEOMETRY ON RULED QUARTIC SURFACES. 51 



turn it about so that it will not contain any other edge of any finite set, 

 and this plane will then cut out three u-edges in addition to the chosen 

 edge ; if /3 is the number of points of C {a) on the chosen edge, we have 

 ^3 4- 3 a = a — k = 4 a, i. e. (3 = a. 



Therefore every edge of the cone has a points of C {a) on it. If the cone 

 has a double or cuspidal edge, an arbitrary plane through it will cut out 

 two a-edges, and if 8 is the number of points of C {a) on the double or 

 cuspidal edge 



8 + 2 a = a — k = 4 a, i. e. 8 = 2 a. 



If the cone has a triple edge, having r points of C ia) on it, an arbitrary 

 plane through this triple edge will cut out one a-edge, and we have 

 t + a = a — k = 4 a, i. e. t = 3 a. 



Therefore, the theorem holds when there is only one infinite set. Sup- 

 pose now that any number of the a sets are infinite, and let a be the 

 least value of p belonging to any of these infinite sets. Pass a cubic 

 cone through nine of these a-edges ; then, since the cubic cone meets 

 C {a) in 3 a points, of which 3 k lie at the vertex and 9 a on the nine 

 chosen edges, the remaining 3 a — 3 k — 9 a points lie on the three 

 other edges in which the cubic and quartic cones intersect, and at least 

 one of these three edges must have as many as a — k — 3 a points of 

 C'- a) on it. Keeping seven of the a-edges fixed, we can vary the other 

 two in such a way that the cubic cone, determined each time by the nine 

 a-edges, will have for its remaining intersection with the quartic cone 

 three edges different from those of any other such cone previously deter- 

 mined ; since there is an infinite number of a-edges, we get, in this way, 

 an infinite number of such cubic cones, and therefore an infinite number 

 of edges each having as many as a — k — 3 a points of C {a) on it ; hav- 

 ing an infinite number of such edges, we can so choose two of tbem that 

 their plane will not pass through any edge of any finite set, because there 

 is a finite number of edges in all the finite sets taken together ; besides 

 the chosen edges this plane will then cut out two edges belonging to the 

 infinite sets. Now the number of points of C {a) on the two chosen edges 

 taken together is equal to or greater than 2 a — 2 k — G a, and if /5 and 

 y be the number of points of C (<t) , respectively, on the other two edges 

 in the plane we must have 



2a — 2k— 6a + j3 + y<«- k, 

 and, since neither (3 nor y can be less than a, 



2a — 2k— 6a + 2a< a — k, 



