We set ? = n M k . Let (n, n ) ? A × B with max(n, n ) ? ? ,
, Theorem 5.15 holds when we restrict ourselves to p = q
, There exists B ? A with positive lower density, min(B) ? a and |n ? n | ? a for all n, Lemma 5.17. Let A ? N with positive lower density and a > 0
, A = {n j : j ? N} in an increasing order and define B = {n ka : k ? N}. We then go inductively from two sets to a sequence of sets
, There exists a sequence (A(p)) of pairwise disjoint subsets of N such that (i) each set A(p) has positive lower density
, ii) for all C > 0, there exists ? > 0 such that, for all (n, n ) ? A(p) × A(q) with p = q and max(n, n ) ? ?
, We shall construct by induction two sequences of sets (A(p)) and (B(p)) and a sequence of integers (? k ) such that, at each step r, (a) for all 1 ? p ? r, A(p) and B(p) are disjoint and have positive lower density
A(q) ? B(p) and B(q) ? B(p) ,
, A(p) and n ? B(q), max(n, n ) ? ? =? |n ? n | ? C
Condition (c) is only helpful for the induction hypothesis. We initialize the construction by applying Lemma 5.16 to E = N. We set A(1) = A and B(1) = B which satisfy (a), (b) and (c). In particular, applying (c) now that the construction has been done until step r and let us perform it for step r + 1. Let E be a subset of B(r) with positive lower density and |n ? n | ? r + 1 provided n = n are in E. We apply Lemma 5.16 to this set E and we set A(r + 1) = A and B(r + 1) = B, so that (a) and (b) are clearly satisfied, r}, for all 1 ? p ? q ? r, for all n ? A(p) and n ? B(q), vol.18 ,
, The proof of (d) is slightly more delicate. For k = 1, . . . , r, we have to verify that for 1 ? p ? r + 1, n ? A(p) and n ? B(r + 1), max(n, n ) ? ? k =? |n ? n | ? k
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