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Source. T. Crmarić and V. Kovač, On the irrationality of certain super-polynomially decaying series, Colloquium Mathematicum (2025), doi:10.4064/cm9628-5-2025; arXiv:2504.18712v1 (25 April 2025). Lemma 4 on p. 4 of the arXiv v1 PDF; its proof on pp. 4--6; Remarks 5 and 6 on p. 6. Bibliographic details and reading limits are in the source card.

Statement

Let $X_1,X_2,X_3,\dots$ be finite subsets of $[0,\infty)$, each with at least two elements, such that $\sum_n\max X_n$ converges. For $n\in\mathbb N$ let $\Delta_n$ and $\delta_n$ be the largest and the smallest length of the intervals into which the points of $X_n$ cut $[\min X_n,\max X_n]$, and put

$$r_n:=\sum_{k=n+1}^{\infty}\bigl(\max X_k-\min X_k\bigr).$$

Consider the set

$$\Bigl\{\ \sum_{n=1}^{\infty}x_n\ :\ x_n\in X_n\text{ for every }n\in\mathbb N\Bigr\} \tag{2.5}$$

• (a) If $r_n\ge\Delta_n$ for every sufficiently large $n$, then (2.5) is a finite union of nondegenerate bounded closed intervals. If $r_n\ge\Delta_n$ for every $n\in\mathbb N$, then (2.5) is the single interval $\bigl[\sum_n\min X_n,\ \sum_n\max X_n\bigr]$ (the paper's (2.6)).
• (b) If $r_n<\delta_n$ for every sufficiently large $n$, then (2.5) is a closed set with empty interior.

Taking $X_n=\{0,x_n\}$ recovers Kakeya's Lemma 3 (p. 3) on the subsums of a convergent series of positive terms. Remark 5 (p. 6) notes that the lemma is already useful under the stronger hypothesis $\max X_{n+1}-\min X_{n+1}\ge\Delta_n$ for all $n$, which is the form used in the proof of Theorem 1. Remark 6 (p. 6) adds that when $r_n<\delta_n$ for every $n$ the measure of (2.5) is $\lim_{N\to\infty}|X_1|\cdots|X_N|\,r_N$.

Proof sketch (pp. 4--6)

For (a) with the hypothesis at every index, a point $x$ of the interval (2.6) is reached greedily: the condition $\Delta_{N+1}\le r_{N+1}$ (the total spread of the sets after $X_{N+1}$) lets each step pick $x_{N+1}\in X_{N+1}$ so that the remainder stays in the interval spanned by the minimal and maximal tails, and these tails tend to $0$. When the hypothesis holds only beyond an index $m$, the set is a finite set of initial sums plus one such interval. For (b), the $N$-th stage cover by intervals of length $r_N$, one per choice of $x_1,\dots,x_N$, consists of pairwise disjoint intervals because $r_l<\delta_l$; since $r_N\to0$ the set has empty interior and, as an intersection of finite unions of closed intervals, is closed; a finite initial segment is handled by the Baire category theorem.

This sketch is written from a reading of the proof's structure; it was not checked line by line.

Read depth. Claims checked: the statement was read clause by clause on p. 4 of the arXiv v1 PDF.

Dependencies

None beyond the convergence of $\sum_n\max X_n$ and, in (b), the Baire category theorem.

Bears on

• Problem 270: the tool behind Theorem 1's negative answer; the lemma by itself settles nothing about the problem.