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Statement

Property QP is defined on the definitions page.

Theorem 2 (p. 5). The following two statements are equivalent.

  1. If AA is any set of positive integers with ∑a∈A1/a=∞\sum_{a\in A}1/a=\infty, then AA has property AP. (The paper labels this statement Erdős' conjecture.)
  2. If AA is any set of positive integers with ∑a∈A1/a=∞\sum_{a\in A}1/a=\infty, then AA has property QP.

Remark in Section 5 (p. 12). The paper says it would be nice to prove Theorem 2 with QP replaced by CP, or Theorem 3 with C replaced by CP, and that proving both is very unlikely, since together they would give Erdős' conjecture.

Source. Brown, T. C., Erdős, P. and Freedman, A. R., Quasi-progressions and descending waves, J. Combin. Theory Ser. A 53 (1990), no. 1, 81--95, doi:10.1016/0097-3165(90)90021-N, read in the authors' copy identified on the source card, whose pages are numbered 1 to 13: the statement and proof on p. 5, the remark on p. 12.

Read depth. Claims checked: the statement and the remark were read clause by clause on the print's pages. The proof was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Section 3, p. 5. Statement 1 implies statement 2 because AP⇒QPAP\Rightarrow QP. For the converse, take AA with infinite reciprocal sum and no kk-term arithmetic progression for a fixed kk. A dilate-translate nA+gnA+g then contains no k−QP(n−1)k-QP(n-1). Choose finite sets Bn⊆nA+gB_n\subseteq nA+g of reciprocal sum greater than 1, with g≥3max⁡Bn−1g\ge3\max B_{n-1}, and let BB be their union. BB has infinite reciprocal sum, and the rapid growth between blocks forces any long QP(d)QP(d) in BB to have kk consecutive terms inside a single BnB_n with n≥d+1n\ge d+1, which is impossible.

Dependencies

The implication AP⇒QPAP\Rightarrow QP of Theorem 1.

Bears on

  • Problem 3: the problem's question is the paper's statement 1, and the theorem shows it equivalent to statement 2, the same assertion with property QP in place of arbitrarily long arithmetic progressions. The equivalence by itself proves neither statement.