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Source. Lemma 4, arXiv:2406.07218v3, PDF p. 5 (the sets $X_{s,t}$ are defined on the same page); proof pp. 5--6.

Dependencies. Lemma 3.

Used in. Theorem 1.

Statement

Write $H_s=\sum_{k=1}^s1/k$ and $|A|$ for the Lebesgue measure of a measurable $A\subseteq\mathbb R$. For integers $0\le s<t$, the set $X_{s,t}$ consists of the $x\in(0,H_s]$ for which there are positive integers $m_1<m_2<\cdots<m_t$ such that $\sum_{k=1}^n1/m_k$ is the best $n$-term Egyptian underapproximation of $x$ for every $n=s,s+1,\dots,t$ (p. 5; the terms are defined on the Theorem 1 page).

Lemma 4 (p. 5): for all integers $100\le s<t$,

$$|X_{s,t+2}|\le\frac{1999}{2000}|X_{s,t}|. \tag{3.3}$$

The set of positive reals with eventually greedy best Egyptian underapproximations is contained in $\bigcup_{s\ge0}\bigcap_{t>s}X_{s,t}$ (the paper's (3.2), p. 5), which is how the lemma feeds Theorem 1.

Proof sketch (pp. 5--6)

A sketch written here. $X_{s,t}$ is a union of classes $(q,r]$ of the partition of $(0,\infty)$ by best $t$-term underapproximation, each of length below $10^{-4}$ since $t>100$. Each class is cut into pieces of the form $q+(1/i,1/(i-1)]$ plus one short leftover piece of relative length below $10^{-4}$. On a piece, a point that stays nested two more steps has $x-q$ with a greedy best two-term underapproximation, so Lemma 3 removes at least one thousandth of the piece; summing over pieces and classes gives the factor $1999/2000$. The proof was read for structure only and has not been independently reviewed.

Bears on. #206: a step in the proof of Theorem 1, the result that answers the problem.