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Statement

Use the notation of Lemma 6. Suppose $N$ is sufficiently large, $N\ge M\ge N^{1/2}$, $\alpha>2(\log N)^{-1/200}$, and $A\subseteq[M,N]\cap\mathbb N$ satisfies

$$R(A)\ge\alpha+(\log N)^{-1/200},\qquad q\le\frac{M}{(\log N)^{1/100}}\quad(q\in\mathcal Q_A).$$

Then some $B\subseteq A$ satisfies

$$\alpha-\frac1M\le R(B)<\alpha,\qquad R(B;q)\ge(\log N)^{-1/100}\quad(q\in\mathcal Q_B).$$

Source. Bloom, arXiv:2112.03726v2, Lemma 7, p. 18.

Rewritten proof

Lemma 6 first gives $A'\subseteq A$ with $R(A')\ge\alpha$ and all nonempty fibers of weight at least $2(\log N)^{-1/100}$. We show how to delete a single element from any current $D\subseteq A'$ with $R(D)\ge\alpha$ and fiber weights at least $(\log N)^{-1/100}$, while retaining that latter bound.

Apply Lemma 6 again, now to $D$, to obtain $C\subseteq D$ with

$$R(C)\ge R(D)-(\log N)^{-1/200} >(\log N)^{-1/200}>0,$$

and $R(C;q)\ge2(\log N)^{-1/100}$ whenever $C_q$ is nonempty. Choose $x\in C$. A fiber not containing $x$ is unchanged. For any fiber containing $x$, also $x\in C_q$, and

$$R(D\setminus\{x\};q) \ge R(C;q)-q/x \ge2(\log N)^{-1/100}-q/M \ge(\log N)^{-1/100}.$$

Repeat this one-element deletion while the mass is at least $\alpha$. The set is finite, so eventually its mass crosses below $\alpha$; it cannot disappear before crossing. Each decrement is $1/x\le1/M$, so the first set below $\alpha$ has mass at least $\alpha-1/M$. All its surviving fiber bounds were preserved at each step.

Dependencies

Lemma 6.

Bears on

• Problem 298
• Problem 299