Ill 



Jordan curve belonging entirely to E. Let P be a third point of J 

 and c an arbitrary circle round P. Now the "Unbewalltheit" says 

 that if Q and R are chosen close enough to P, the joining curves 

 may be kept entirely inside c. 



Applying this to our case it follows that points of ^^6', and J^D^ 

 can be joined by open Joijuan curves entirely belonging to Ij, inside 

 any vicinity of A^. Hence in the continuous^ (1,1) representation 

 AC and AD can be joined by open Jordan curves entirely situated 

 on I inside any vicinity of .4. Now every one of these curves 

 has at least one point in common with /^, because AC and AD lie 

 on ditferent sides of that plane, hence in plane ,i the point A is 

 limiting point of 1, and in the same way can be proved that A is 

 limiting point of III in /i. But I and III have ]io points in common, 

 hence in /^ one bi-anch departs from A on 1 and another on 111. I and 111 

 are both situated above «so in /? two branches depart from yl above ^'. 



In [3 two branches arrive at ^ from the same side of 6. Considering 

 the possible forms of elementary curves of the third order, there 

 are a priori three possibilities: 



1. ^ is double point in [3. 



2. A is cusp in /?. 



3. A is ordinary point in /? with b for tangent. 



1. Suppose A is double point in /I Two branches AP and AQ 

 arrive in A from above b, hence two more AS and AR arrive from 

 below b (three from one side and one from the other is impossible 

 because b has, besides A, another point in common with F'*): We 

 proceed to show that the branches AR and AS are at first both 

 situated on II or both on IV. Suppose AR and AS were situated 

 respectively on II and IV. Then AR and ^*S could not be connected 

 below «, because II and IV have no points in common. But AS 

 would be connected via AC and AF with AP and AQ and AR 

 would be connected via AD and AF with AP and AQ. From this 

 follows that AR and AS would only be connected vm AP and AQ. 

 This however leads to a contradiction, because the four branches 

 meeting at A in /? must be connected in an analogous way as those 

 in « hence AR and ^46' are joined by a set of points situated 

 entirely on one side of [i. Thus it has been shown that AR and AS 

 are situated either both on II or both on IV, let us assume on II. 



The vicinity of A on F^ is the (1,1) continuous representation 

 of the vicinity of a point in a plane. Let A correspond to A^, 

 AF to A,F„ AD to A,D, II, to 11^, AR to A,R, and AS to A,S,. 

 Inside a finite neighbourhood of A^ the region U^ is divided by the 



