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By S. Payne, T. Thas
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L [ M / D ¿. Evidently we may also assume that fx; yg?? L [ M / D ¿. Let x 2 fx1 ; x2 g?? , y 2 fy1 ; y2 g?? , y1 2 L and y2 2 M ). First, we suppose that fx1 ; x2 g?? , x1 6 x2 ). Then let z1 and N be defined by x2 L . Since we have proved that the union P 00 of x I N I z1 I L. Clearly N ?? the sets fz; ug , z 2 N and u 2 M , coincides with P 0 , the point y belongs to P 00 . By a preceding case, fx; yg?? P 00 , so fx; yg?? P 0 . Second, we suppose x1 x2 , and without loss of generality that y1 y2 .
The incidence I is the natural one. q 1; q C 1/. qC2/, and that each element of B is incident with q elements of P . For each value of c there are q 2 curves of type (iii), and these curves have no point in common. a0 ; b 0 ; c 0 / intersect. Then for some we have c 2 b C a D c 2 b 0 C a0 and 2c C b 0 D 2c C b, which clearly implies b D b 0 and a D a0 . Similarly, no two curves of the form (i) (or of the form (ii)) intersect. Thus we have q C 2 families of nonintersecting curves, q 2 curves in each family and q points on each curve.
As y I x Â , we have x I y Â . Since y I y Â , clearly y Â D xy D L D x Â , implying x D y, a contradiction. So each line of S is incident with at most one absolute point of Â . A line L is absolute iff L I LÂ iff LÂ is absolute. , L I LÂ . If LÂ I M I u I L, then LÂ I uÂ I M Â I L, hence uÂ D M and M Â D u. Consequently u and M are absolute, and we have proved that each line L is incident with at least one absolute point. It follows that the set of absolute points of Â is an ovoid. Dually, the set of all absolute lines is a spread.