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Source. S. Fan, Strongly complete sets and a conjecture of Erdős, arXiv:2607.14071v5 (16 September 2026): Remark 4.1, the case $\rho=2$, physical and printed pp. 19--20 (the base-two example already opens Remark 4.1 of v4, p. 19, as the bound $M_2^*\ge2$, with the same set, the same displayed inequality and the same incompleteness count; v5 generalizes the remark to $M_\rho^*\ge u_\rho$ for $\rho\ge2$, adds the case $\rho>2$ and the random-set sentence, and moves the remark's second paragraph to Remark 4.2). Read in the canonical conversion beside the held v5 PDF; the artifact is identified on the library source card, Fan (2026). The remark has no result page of its own on the card; its consequence $2\le M_2^*\le5$ is recorded on the card's Remark 4.2 page. The source prints the per-interval count as $|A\cap[2^k,2^{k+1}]|=u_\rho=1$ (v5 p. 19); since $u_2=\lceil2(2-1)\rceil=2$ by (1.6) on p. 3, the middle term is a slip, and the count is read as $u_\rho-1=1$, the count the bound $M_\rho^*\ge u_\rho$ needs (v4 prints $=1$).

Standing. This is an author-recorded reconstruction. It is not an independent review, changes no status and assigns no tier. Only the case $\rho=2$ is reconstructed; the source's case $\rho>2$ (giving $M_\rho^*\ge u_\rho$) and its unproved-here statement about random sets are omitted.

Definitions

$\mathbb N=\{1,2,\ldots\}$, $\operatorname{FS}$, complete, strongly complete, condition (1.5) and $M_\rho^*$ are as on the Remark 4.2 page.

Statement

Remark 4.1, base two. The set $A=\{2^k+1:k\in\mathbb N\}$ satisfies (1.5), has exactly one element in every $(2^k,2^{k+1}]$ with $k\ge1$, and is not complete. Hence $M_2^*\ge2$, and with Corollary 1.2, $2\le M_2^*\le5$.

Proof

One element per interval. For $k\ge1$, $2^k+1\in(2^k,2^{k+1}]$, and $2^j+1$ for $j\ne k$ lies outside this interval.

Condition (1.5). Let $\theta\in\mathbb R\setminus\mathbb Z$ with $\sum_{a\in A}\|a\theta\|<\infty$. Then $\|(2^k+1)\theta\|\to0$ and $\|(2^{k+1}+1)\theta\|\to0$ as $k\to\infty$, and

$$\|\theta\|=\|2(2^k+1)\theta-(2^{k+1}+1)\theta\| \le2\|(2^k+1)\theta\|+\|(2^{k+1}+1)\theta\|\to0,$$

so $\|\theta\|=0$, a contradiction.

Incompleteness. For $k\ge1$, the elements of $A$ that are at most $2^{k+1}$ are $2^j+1$ for $1\le j\le k$, so $|A\cap[1,2^{k+1}]|=k$. Every element of $\operatorname{FS}(A)\cap[1,2^{k+1}]$ is a sum of a nonempty subset of these $k$ elements, so $|\operatorname{FS}(A)\cap[1,2^{k+1}]|\le2^k-1$, leaving at least $2^{k+1}-(2^k-1)=2^k+1$ integers of $[1,2^{k+1}]$ outside $\operatorname{FS}(A)$. As $k\to\infty$ this count is unbounded, so $\mathbb N\setminus\operatorname{FS}(A)$ is infinite and $A$ is not complete (nor, a fortiori, strongly complete).

Conclusion. $A$ satisfies (1.5) and has at least one element in every large dyadic interval but is not strongly complete, so the threshold $M_2^*$ exceeds $1$. Corollary 1.2 of the source (statement on the Remark 4.2 page, proof not reconstructed) gives $M_2^*\le5$.

Scope. The bound $M_2^*\ge2$ is the only part of Remark 4.1 reconstructed here. The exact value of $M_2^*$ is open in the source; its relevance to Problem 354 is through Remark 4.2.