116 



a closed set of points to which A does not belong, none of the lines 

 Ci,c,,c, . . . . b being tangent. 



Let e^, e^, e^ . . . be the lines of intersection of i^ and y,, y^, y, . . . 

 respectively. These lines e.^^, e^, e^ . . . converge towards e. 



In y, a branch departs from A between c^ and e^, in y, between 

 c, and e^ etc. The distance from A at which these branches can 

 cross Cj, Cj.Cj. . . . cannot tend towards zero, hence to make it possible 

 that in plane /? no branch departs from A between h Siixd e it is 

 unavoidable that the branches in the converging planes cross 

 ^1, e^, e^ . . . in points converging towards A. This means that in plane 

 (^ the line e would be tangent in A. But considering d does not 

 coincide with b or either of the tangents in «, the plane (^ through 

 d must show an ordinary point in A with d for tangent. Thus a 

 contradiction has been obtained. 



It has been shown successively that the a priori possibilities 1 and 

 2 given on page HI lead to contradictory results. Hence only the 

 third possibility remains, namely that A is ordinary point in /J with 

 b for tangent. But b is an arbitrary line in <( through A, only 

 subjected to the condition not to coincide with either of the tangents 

 in A, and ,? is an arbitrary plane through b, only supposed not to 

 coincide with ((, hence the results obtained so far may be expressed 

 as follows : In every plane through A luJdch does not coincide with 

 a and does not contain a tangent in «, the point A is ordinary point 

 lüith tangent situated in ft. 



Thus to complete the proof that a is tangent plane, it only remains 

 to consider the sections of F^ in planes through a tangent at .4 in «. 



In (( the point A is point of intersection of two convex arches, 

 parts of which are indicated by QS and PR in fig. 5. Let r/ (^ Z)C) 

 be tangent at A to PR, and let /? be an arbitrary plane through 

 a {—\=a). We assume the senses of curvature of the convex arches 

 to be as indicated in fig. 5. 



In ^ we choose a line AB{^~a) and we consider a sequence of 



planes /Jj, /?,, /?, all passing through AB and converging towards ^, 



in such a way that the back part converges towards ^ from the right 

 hand side (see fig. 5). The line of intersection of « and /?„ is denoted 

 by ACn {a„). 



Let the part of F^ connecting AP and AS be situated above « 

 (the other case is treated in a strictly analogous way). In every 

 plane ^„ a branch departs from A above « in the direction ACn. 

 These branches have a limiting set in /? belonging to the closed set 

 F^. Applying the same reasoning given above to show that J cannot 

 be cusp in any plane, it can be shown that this limiting branch 



