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second order, but at the same time have 3 different points in common 
with a line and this of course cannot be. 
Third case. Let a be the plane, b and c the lines and A their 
point of intersection. If we exclude the point A, then along the rest 
of 6 the sectors of /'* meet either everywhere from the same side, 
or always from opposite sides of a’). The same holds for c. Now 
if the sectors meet from the same side of « both along 6 and c, 
then an infinite number of lines of a is contained in /’*. If the 
sectors meet from the same side along b and from opposite sides 
along c, then, according to our definitions, 6 counts double in «, 
hence the section would be of the third order. Remains the possi- 
bility that along 6 as well as c the sectors meet from different sides 
of a. Let d be an arbitrary plane, not containing A, a the line of 
intersection of d and @ and B and C the points of intersection of 
a and ce resp. 6. The curve in d crosses a at B and C and has 
no further points in common with a. For a curve of the third order 
this would imply that either B or Cis double point and the reasoning 
used for the second case reduces this to a contradiction. Hence the 
curve in d is evidently of the second order. On the other hand there 
exist lines not passing through A and carrying 3 different points of 
F*. Thus in a plane through one of these lines and not containing 
A we have once more constructed an impossibility. 
Fourth case. Let a be the plane and a the double line. Along a 
the sectors of F'? meet either everywhere from the same side, or 
always from opposite sides of a. The former is impossible as the 
space in which we work is supposed to be projective, and the 
latter would imply that through every line (4 a) of a a plane 8 
passes with a double point on a. This obviously would mean two 
lines in 8. Hence /#'* would contain an infinite number of lines, a 
possibility excluded throughout. 
This completes our demonstration. 
From the preceding theorem follows that a plane « with a double 
line a contains besides a single line 6. Let a and 6 intersect at A. 
We proceed to show that along a (point A excluded) the sectors of 
1) Every point of b (4 A) is internal point of an interval along which the 
sectors meet either from the same or from opposite sides of « Excluding an 
arbitrarily small open segment of b round 4A, a finite number of these intervals 
exist, such that every point of the rest of b is internal to at least one of them 
(Borer-theorem). From this the property under consideration follows at once for 
the entire line & (excluded point A). 
