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Source. S. Fan, Strongly complete sets and a conjecture of Erdős, arXiv:2607.14071v5 (16 September 2026): Remark 4.2, physical and printed p. 20 (Remark 4.1 of v4, p. 19, with the same content); the definitions (1.1), (1.5), (1.8), (1.9), the observation that strongly complete sets satisfy (1.5), Theorem 1.1 and Corollary 1.2, pp. 2--4. Read in the canonical conversion beside the held v5 PDF, which was not itself opened for this page; the artifacts are identified on the library source card, Fan (2026), and the remark on its result page. The remark's rescaling, the finiteness of the intersection of the two rays and the "routine" triangle-inequality step are stated without detail in the source; they are written out below.

Standing. This is an author-recorded reconstruction. It is not an independent review, changes no status and assigns no tier. Theorem 1.1 and Corollary 1.2 are used only as statements: their proofs (Sections 2--4 of the source, about twenty pages) are not reconstructed.

Definitions

Here $\mathbb N=\{1,2,\ldots\}$, as in the source. For $A\subseteq\mathbb N$, $\operatorname{FS}(A)$ is the set of sums of nonempty finite subsets of $A$; $A$ is complete if $\mathbb N\setminus\operatorname{FS}(A)$ is finite and strongly complete if $A\setminus B$ is complete for every finite $B\subseteq A$. $\|x\|$ is the distance from the real $x$ to the nearest integer. Condition (1.5) for $A$ is

$$\sum_{a\in A}\|a\theta\|=\infty \qquad\text{for every }\theta\in\mathbb R\setminus\mathbb Z .$$

$M_\rho^*$ (1.8) is the least positive integer such that every $A\subseteq\mathbb N$ satisfying (1.5) with $|A\cap(\rho^k,\rho^{k+1}]|\ge M_\rho^*$ for every sufficiently large $k$ is strongly complete. For $\alpha,\beta>0$,

$$A_{\alpha,\beta}=\{\lfloor2^k\alpha\rfloor,\lfloor2^k\beta\rfloor:k\ge0\} \setminus\{0\}\qquad(1.9);$$

$\alpha\sim\beta$ means $\alpha/\beta=2^n$ for some $n\in\mathbb Z$, and $\alpha$ is a dyadic rational if $\alpha\sim n$ for some nonzero integer $n$. For $x>0$ and $K\ge0$ let $U_K(x)=\{\lfloor2^kx\rfloor:k\ge K\}$. Hegyvári's conjecture, as the source records it (p. 4): $A_{\alpha,\beta}$ is complete whenever $\alpha\not\sim\beta$ and at least one of $\alpha,\beta$ is not a dyadic rational.

In-source theorems used as statements (not reconstructed). Theorem 1.1 (p. 3): for $\rho>1$, with $u_\rho=\lceil\rho(\rho-1)\rceil$, $v_\rho=\lceil\rho^3/(\rho+1)\rceil$ and $M_\rho=\min\{2u_\rho+1,2v_\rho\}$, every $A\subseteq\mathbb N$ satisfying (1.5) and $|A\cap(\rho^k,\rho^{k+1}]|\ge M\ge M_\rho$ for all large $k$ has $q_{A\setminus F}(n)/n^{(M-M_\rho)\log_\rho2}\to\infty$ for every finite $F\subseteq A$, where $q_B(n)$ counts representations of $n$ as sums of distinct elements of $B$; in particular $A$ is strongly complete. Corollary 1.2 (p. 4), the case $\rho=2$ where $u_2=2$, $v_2=3$, $M_2=5$: every $A$ satisfying (1.5) with at least five elements in $(2^k,2^{k+1}]$ for all large $k$ is strongly complete, so $M_2^*\le5$.

Statement

Observation (p. 3). Every strongly complete $A\subseteq\mathbb N$ satisfies (1.5).

Remark 4.2. If $M_2^*=2$, then $A_{\alpha,\beta}$ is strongly complete whenever $\alpha\not\sim\beta$ and at least one of $\alpha,\beta$ is not a dyadic rational. In particular $M_2^*=2$ would imply Hegyvári's conjecture, in the stronger form of strong completeness.

Proof

The observation

Suppose $\sum_{a\in A}\|a\theta\|<\infty$ for some $\theta\in\mathbb R\setminus\mathbb Z$, so $\|\theta\|>0$. Choose $N_0$ with $\sum_{a\in A,\,a>N_0}\|a\theta\|<\|\theta\|/2$. Since $A$ is strongly complete, $A\cap(N_0,\infty)$ is complete, so every sufficiently large $n$ and $n+1$ are sums of distinct elements of $A\cap(N_0,\infty)$, and by the triangle inequality on $\mathbb R/\mathbb Z$ each of $\|n\theta\|$, $\|(n+1)\theta\|$ is at most the sum of $\|a\theta\|$ over the elements used, hence less than $\|\theta\|/2$. Then $\|\theta\|=\|(n+1)\theta-n\theta\|\le\|(n+1)\theta\|+\|n\theta\|<\|\theta\|$, a contradiction.

Step 1: rescaling

Assume, by symmetry, that $\alpha$ is not a dyadic rational. Let $s,t$ be the integers with $\alpha'=2^{-s}\alpha\in(1/2,1]$ and $\beta'=2^{-t}\beta\in(1/2,1]$. Then $\alpha'\ne\beta'$ (else $\alpha/\beta=2^{s-t}$), and $\alpha'$ is not a dyadic rational, since $\alpha'\sim\alpha$. The set $U_0(\alpha)=\{\lfloor2^{k+s}\alpha'\rfloor:k\ge0\}$ contains $U_{k_0}(\alpha')$ for every $k_0\ge\max(s,0)$, and similarly for $\beta$; so for $k_0\ge\max(s,t,1)$,

$$U_{k_0}(\alpha')\cup U_{k_0}(\beta')\subseteq A_{\alpha,\beta}$$

(the values are positive, as $2^{k_0}\alpha'>1/2\cdot2^{k_0}\ge1$ for $k_0\ge1$), and the complement $B=A_{\alpha,\beta}\setminus(U_{k_0}(\alpha')\cup U_{k_0}(\beta'))$ is finite: an element $\lfloor2^k\alpha\rfloor$ of $A_{\alpha,\beta}$ with $k+s\ge k_0$ lies in $U_{k_0}(\alpha')$, so $B$ consists of values with $k<k_0-s$ or $k<k_0-t$.

Finiteness of $U_0(\alpha')\cap U_0(\beta')$. For $k\ge1$, $2^k\alpha'\in(2^{k-1},2^k]$, so $\lfloor2^k\alpha'\rfloor\in[2^{k-1},2^k]$, and likewise for $\beta'$. If $\lfloor2^k\alpha'\rfloor=\lfloor2^j\beta'\rfloor$ with $k,j\ge1$, the two ranges $[2^{k-1},2^k]$ and $[2^{j-1},2^j]$ must meet, so $|k-j|\le1$. The case $j=k+1$ forces the common value to be $2^k$, so $\lfloor2^{k+1}\beta'\rfloor=2^k$, that is $\beta'<1/2+2^{-k-1}$, which fails for $k$ large since $\beta'>1/2$; $j=k-1$ is excluded symmetrically for $k$ large; and $j=k$ fails for $k$ large since $2^k|\alpha'-\beta'|\ge2$ then makes the floors differ. So only finitely many coincidences occur, and for $k_0$ large $U_{k_0}(\alpha')\cap U_{k_0}(\beta')=\emptyset$. From now on $k_0$ is large enough for this and for $k_0\ge\max(s,t,1)$; the union above with $B$ is then a partition of $A_{\alpha,\beta}$.

Step 2: two elements in every large dyadic interval

For $k\ge k_0$, $2^{k+1}\alpha'\in(2^k,2^{k+1}]$, so $\lfloor2^{k+1}\alpha'\rfloor\in[2^k,2^{k+1}]$, and it equals $2^k$ only if $\alpha'<1/2+2^{-k-1}$, which fails for $k$ large. Hence for $k_0$ large and $k\ge k_0$ both $\lfloor2^{k+1}\alpha'\rfloor$ and $\lfloor2^{k+1}\beta'\rfloor$ lie in $(2^k,2^{k+1}]$; they are distinct by Step 1 and belong to $A_{\alpha,\beta}$. Thus

$$|A_{\alpha,\beta}\cap(2^k,2^{k+1}]|\ge2\qquad(k\ge k_0).$$

Step 3: condition (1.5)

Write $c_k=\lfloor2^k\alpha'\rfloor$ and $d_k=c_{k+1}-2c_k\in\{0,1\}$ (as $\lfloor2x\rfloor-2\lfloor x\rfloor\in\{0,1\}$). If $d_k=0$ for all $k\ge k_1$, then $c_k=2^{k-k_1}c_{k_1}$ for $k\ge k_1$, and $2^{-k}c_k\to\alpha'$ gives $\alpha'=c_{k_1}/2^{k_1}$, so $\alpha=2^{s-k_1}c_{k_1}$ with $c_{k_1}$ a nonzero integer, contradicting that $\alpha$ is not a dyadic rational. Hence $c_{k+1}=2c_k+1$ for infinitely many $k$.

Let $\theta\in\mathbb R\setminus\mathbb Z$ with $\sum_{a\in A_{\alpha,\beta}}\|a\theta\|<\infty$. The elements $c_k$, $k\ge k_0$, belong to $A_{\alpha,\beta}$ and are pairwise distinct, so $\|c_k\theta\|\to0$. For the infinitely many $k$ with $c_{k+1}=2c_k+1$,

$$\|\theta\|=\|c_{k+1}\theta-2c_k\theta\|\le\|c_{k+1}\theta\|+2\|c_k\theta\|\to0,$$

so $\|\theta\|=0$, contradicting $\theta\notin\mathbb Z$. Hence $A_{\alpha,\beta}$ satisfies (1.5).

Step 4: conclusion

By Steps 2 and 3, $A_{\alpha,\beta}$ satisfies (1.5) and has at least two elements in every $(2^k,2^{k+1}]$ with $k\ge k_0$. If $M_2^*=2$, the definition of $M_2^*$ makes $A_{\alpha,\beta}$ strongly complete. This is the remark.

Scope. The hypothesis $M_2^*=2$ is unproved: the source proves $M_2^*\le5$ (Corollary 1.2) and $M_2^*\ge2$ (Remark 4.1, reconstructed on the next page). Conversely, strong completeness of every $A_{\alpha,\beta}$ under Hegyvári's condition would not by itself give $M_2^*=2$, since these sets are special. Nothing here concerns bases other than $2$, and the argument is silent on the rational-ratio cases beyond showing that they would follow from the sharp threshold.