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Statement

Let AA be a set of natural numbers and BB the set of natural numbers divisible by no element of AA.

Lemma 2.1 (printed p. 388). For all yy,

∑b≤yb∈B1b ≥ ∏a∈A(1−1a)log⁡(y+1)\sum_{\substack{b\le y\\ b\in B}}\frac1b \ \ge\ \prod_{a\in A}\Bigl(1-\frac1a\Bigr)\log(y+1)

(display (2.2)).

The lemma sits in the proof of Theorem 2, where AA is the set of that theorem; its statement and proof use no further hypothesis on AA. If 1∈A1\in A both sides vanish. A note after the proof says that the lemma gives, as a by-product, a proof of the Heilbronn–Rohrbach inequality (1.6), p. 387: for a fixed AA, the density lim⁡x→∞F(x,A)/x\lim_{x\to\infty}F(x,A)/x of the integers divisible by no element of AA is at least ∏a∈A(1−1/a)\prod_{a\in A}(1-1/a).

Source. P. Erdős and I. Z. Ruzsa, On the small sieve. I. Sifting by primes, J. Number Theory 12 (1980), 385–394; Lemma 2.1 on printed p. 388 (PDF p. 4), with the inequality (1.6) on p. 387. The edition is identified in the source digest.

Read depth. Claims checked: the statement and the note were read on the page images. The proof was not checked.

Proof pointer

Every natural number factors as a product of powers of elements of AA times an element of BB, so the harmonic sum up to yy is at most the sum of 1/b1/b over b∈Bb\in B, b≤yb\le y, times ∏a∈A(1−1/a)−1\prod_{a\in A}(1-1/a)^{-1}; comparing the harmonic sum with log⁡(y+1)\log(y+1) gives (2.2).

Dependencies

None.

Bears on

No problem directly. It is the input to Theorem 2, which bears on Problem 784.