---

Source. Liu--Sawhney, On further questions regarding unit fractions, arXiv:2404.07113v1, Lemma 2.4, p. 8; see the source digest.

Statement. Fix a sufficiently large $X$ (the print leaves this largeness implicit) and an integer interval $I$ of length $X$. Suppose $\mathcal P$ is a set of primes none of which exceeds

$$z=\exp\!\left(\frac{\log X}{\sqrt{\log\log X}}\right).$$

Put $I_p=\{n\in I:p\mid n\}$ and $R(\mathcal P)=\sum_{p\in\mathcal P}1/p$. Then, with absolute comparison constants,

$$\left|I\setminus\bigcup_{p\in\mathcal P}I_p\right| \asymp X\prod_{p\in\mathcal P}\left(1-\frac1p\right) \asymp X\exp(-R(\mathcal P)).$$

External input. The fundamental lemma of sieve theory in Koukoulopoulos, The Distribution of Prime Numbers, Graduate Studies in Mathematics 203, AMS (2019), Theorem 18.11(b), is the cited input. The author-hosted preliminary version has Theorem 18.11(b) on printed p. 190/PDF p. 201. The paper invokes its dimension-one case with sieve level $D=X^{1/2}$ and parameter $m=1$. The external theorem itself is not proved here.

Proof. Write $P=\prod_{p\in\mathcal P}p$. For each $d\mid P$, counting multiples in an interval gives

$$|\{n\in I:d\mid n\}|=\frac Xd+r_d, \qquad |r_d|\leq2.$$

This is the first sieve axiom in the cited theorem, with density $\nu(d)=1$. The second, dimension-one sieve axiom holds with an absolute constant: products of $(1-1/p)^{-1}$ over subsets of primes are bounded by the corresponding products over all primes, which have the usual dimension-one prime-product bound. The third axiom, controlling the weighted remainder sum at level $D$, follows from

$$\sum_{d\leq D}\tau(d)|r_d| \leq2\sum_{d\leq D}\tau(d) \ll D^{1+o(1)}=X^{1/2+o(1)}.$$

Here $\tau(d)$ is the divisor function; the inequality follows, for example, by counting pairs of positive integers with product at most $D$. The ratio of the logarithm of the sieve level to the logarithm of the largest allowed sieving prime is

$$\frac{\log D}{\log z}=\frac12\sqrt{\log\log X}\longrightarrow\infty.$$

The fundamental lemma therefore gives the first comparison in the statement. These are exactly the parameters and three axiom checks used by the source.

Finally, uniformly for $p\geq2$,

$$\log(1-1/p)=-1/p+O(1/p^2).$$

Since $\sum_p1/p^2<\infty$, summing over $\mathcal P$ and exponentiating gives

$$\prod_{p\in\mathcal P}(1-1/p) =\exp(-R(\mathcal P)+O(1)),$$

which proves the second comparison.

Dependencies. The external fundamental lemma cited above, elementary interval counting, and standard dimension-one prime-product estimates (the paper groups the latter with its number-theoretic preliminaries, Theorem 2.1).

Bears on. #298 and #299, through the sieve estimates in the quantitative reciprocal-sum argument.