306 



Let c be an arbitrary line through A in y, not situated in a or 

 /?. Let the plane y revolve round c. If in any position of y the 

 point A is isolated then our object is attained. The alternative is 

 that ^4 is double point in all planes through c except the plane 

 through c and. a. The foregoing results (p. 113— 114) show that the 

 only way to escape immediate contradiction is to assume A cusp in 

 the plane through c and a. But c was an arbitrary line through A in 

 y only subjected to the condition not to be situated in ^<r or /?, hence 

 every plane through a would show a cusp in A and the reasoning 

 given on page 108 shows that then A would be isolated in every 

 plane not containing a. 



Second assumption : The cuspidal tangents do not coincide. The 

 line of intersection a of the planes a and '^, in which A is cusp, 

 cannot be cuspidal tangent in either of these planes, because a has, 

 except A, another point in common with i'^'. Hence the case indicated 

 in fig. 6 includes all possibilities. Let BEF CD be a [)lane _l a. 



The semiplanes aE and aD contain no points of F^ inside a certain 

 finite neighbourhood of A. On p. 104— 105 it was shown that if .1 is 

 isolated in a plane u, then on one side of « there is a finite 

 neighbourhood of A containing no points of F\ The demonstration 

 was entirely based on the analysis situs, hence it is of no consequence 

 whether the semiplanes in which a is divided by a line through A, 

 happen to make an angle of 180° with each other or any other 

 angle (=|= zero). Applying this to the case of fig. 6 it follows that 

 there exists a finite neighbourhood of A containing no points of 

 F^ inside that part of space situated between the semiplanes aE 

 and aD and in which the semiplanes aF and aC are not situated 

 (in the semiplanes aF and aC branches meet at A, so in this angle 

 between aE and aD the point A is certainly not isolated). 



