741 



one side there even can be three, when the loop reaches the line 

 at infinity, but in any case there is at least, one on either side). 

 Now let (f revolve round c. The curve in a limiting plane is the 

 limiting set of the curves in the converging planes (no isolated points 

 are possible here). Besides a sequence of intinite branches has an 

 infinite limiting branch. Hence it follows that in every plane through 

 c {=\= a) we can choose on both sides of c an infinite branch such 

 that they merge in each other in continuous fashion when d revolves 

 round c. If we add the line a in a these branches are just sutTicient 

 to give F^ the character of a twodimensional continuum in tlie 

 neighbourhood of A and the branches departing from A, which we 

 have left out, cannot be fitted in anymore. This contradicts our 

 assumption that F"^ is a twodimensional continuum (of course the 

 neighbourhood of a point on a twodimensial continuum can in an 

 infinite number of ways be represented on the neighbourhood of a 

 point in a plane, but the neighbourhood of a point in a plane can 

 by (1,1) continuous transformation in the plane never be transformed 

 in anything but the neighbourhood of a point). 



Second case. A is situated on the line a of F^ and is cusp in /J 

 and y. The cuspidal tangents do not coincide, hence the line of 

 intersection 6 of |i and y cannot be cuspidal tangent in either of 

 these planes. It follows that b carries besides A a second point B 

 of F^ and the curve in the plane a through a and b consists of a 

 and an oval through A and B. The line b divides /? in two semi- 

 planes : in the one the cuspidal branches depart from A, hence in 

 the other A is isolated. In the same way A is isolated in one of the 

 semiplanes in which b divides y. 



Now a foregoing demonstration (§ 5, second communication) shows 

 that in this case A is isolated inside the entire angle (<^ 180°) 

 between these semiplanes. Hence the line a belonging to F^ cannot 

 pass through this angle and it follows that the semiplanes of /i and 

 Y in which the cuspidal branches depart from A, are situated on 

 the same side of the plane « through a and b, let us say below «. 

 In a four bi'anches arrive at A, consecutively AP, AQ, AR and 

 AS (two on a and two on the oval). Suppose above ((, AP \^ 

 connected with AQ and AR with AS. Then line b must lie inside 

 the angles QAR and PAS, because planes pass through b in which 

 A is isolated above a. Let c be a line in « through A inside the 

 angles PAQ and SAR fthis is impossible when the oval in « has 

 a for tangent at A, which case we shall consider separately). The fore- 

 going results show that A is double point or cusp in every plane 



53 

 Proceedings Royal Acad. Amsterdam Vol. XX. 



