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Statement

Setting (p. 2). For a finite set $D$ of positive integers, $\mathcal H(D)$ is the hypergraph on the vertex set $D$ whose hyperedges are the sets $\{d\}\cup E\subseteq D$ with $d\notin E$, $E\ne\varnothing$ and

$$\frac1d=\sum_{e\in E}\frac1e$$

(the paper's equation (2)). A subset of $D$ is independent when it contains no hyperedge, and $\alpha(D)$ is the largest size of an independent subset of $D$. A forbidden solution is a choice of pairwise distinct $a,b_1,\ldots,b_k$ with $k\ge1$ and $1/a=1/b_1+\cdots+1/b_k$ (the paper's equation (1), p. 1), and $[N]=\{1,\ldots,N\}$.

Lemma 1 (Dilation lemma, p. 2). Let $D$ be a finite set of positive integers and $m$ a positive integer. If $A\subseteq[N]$ contains no solution of (1), then

$$|A\cap mD|\le\alpha(D)$$

whenever $mD\subseteq[N]$.

Source. Xinjun Wang, A 667/806 Upper Bound for Erdős Problem #301 on Unit-Fraction-Free Sets, unpublished manuscript dated May 27, 2026 on its title page, posted on ResearchGate (2026), identified on the source card: Lemma 1 and the definition of $\mathcal H(D)$ on p. 2, in Section 2 (pp. 2--3). An unrefereed manuscript, which the site's Problem 301 discussion thread describes as AI-generated.

Read depth. Claims checked: the statement and the definitions it uses were read clause by clause on the page image. Nothing here is independently reviewed.

Proof pointer

P. 2. If $A\cap mD$ had more than $\alpha(D)$ elements, the elements of $D$ it comes from would contain a hyperedge $\{d\}\cup E$; multiplying the reciprocal identity through by $1/m$ turns it into a solution of (1) with pairwise distinct terms $md$ and $me$ ($e\in E$), all in $A$.

Dependencies

None.

Bears on

• Problem 301: the lemma is the step that turns a finite certificate on $D$ into a count of elements a relation-free set must omit from each dilate; the manuscript's Theorem 1 applies it to the prefixes of the divisor set of Proposition 1. The lemma alone gives no density bound for $f(N)$.