GPT 6 Astra Ultra attempt on Math-5527
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Contents
Erdős Problem 245: doubling of an infinite set of density zero
1) Statement and scope
For an infinite set $A\subseteq\mathbb N$ with $A(N)=o(N)$, must \[ \limsup_{N\to\infty}\frac{(A+A)(N)}{A(N)}\geq3? \] Here $S(N)=|S\cap[1,N]|$. Freiman proved the affirmative answer. This attempt establishes the sharpness of the constant by a fully specified example with limsup equal to three. The universal lower bound is unresolved in this attempt.2) The block construction and zero density
For $j\geq1$, put \[ x_j=2^{4^j},\qquad \ell_j=\sqrt{x_j},\qquad I_j=[x_j,x_j+\ell_j]\cap\mathbb Z,\qquad A=\bigcup_{j\geq1}I_j. \] Let $B_j=\bigcup_{i<j}I_i$, $b_j=|B_j|$, and $M_j=\max B_j$ for $j\geq2$. Since $x_j=x_{j-1}^4$ and $\ell_j=x_{j-1}^2$, \[ b_j=O(\sqrt{x_{j-1}}),\qquad M_j=O(x_{j-1}),\qquad b_j^2+M_j=o(\ell_j),\qquad b_j=o(\ell_j). \] For $x_j\leq N<x_{j+1}$ we have $A(N)\leq b_j+\ell_j+1=O(\sqrt{x_j})$ and $N\geq x_j$. Therefore $A(N)/N\to0$.3) Bounding the limsup from above
First let $x_j+\ell_j\leq N<x_{j+1}$. Only the first $j$ blocks can contribute to $(A+A)\cap[1,N]$. Old--old sums have at most $b_j^2$ possible values. New--old sums lie in an interval of length at most $\ell_j+M_j+1$. New--new sums lie in $[2x_j,2x_j+2\ell_j]$, of size $2\ell_j+1$. Consequently \[ (A+A)(N)\leq 3\ell_j+b_j^2+M_j+2 =3\ell_j+o(\ell_j), \qquad A(N)=b_j+\ell_j+1. \] The ratio is at most $3+o(1)$ throughout this range.Next let $N=x_j+t$, where $0\leq t\leq\ell_j$. No new--new sum is at most $N$. All old--old sums are already below $x_j$, while the new--old sums at most $N$ lie in an interval of size at most $t+1$. At $x_j-1$, the preceding completed-block bound gives $|B_j+B_j|\leq(3+\varepsilon_j)b_j$, with $\varepsilon_j\to0$. Thus \[ \frac{(A+A)(x_j+t)}{A(x_j+t)} \leq\frac{(3+\varepsilon_j)b_j+t+1}{b_j+t+1} \leq3+|\varepsilon_j|. \] This also covers the partially filled blocks, so the global limsup is at most three.
4) Attaining three
Fix one $a_*\in I_1$. For large $j$, at $N_j=2x_j+2\ell_j<x_{j+1}$, both intervals \[ a_*+I_j,\qquad I_j+I_j=[2x_j,2x_j+2\ell_j]\cap\mathbb Z \] are present and disjoint. Their total size is $3\ell_j+2$. Since $A(N_j)=b_j+\ell_j+1=(1+o(1))\ell_j$, the ratio has limsup at least three, and hence exactly three.For comparison, a much weaker universal bound follows from $a_*+A\subseteq A+A$: $(A+A)(N)\geq A(N-a_*)\geq A(N)-a_*$, so the liminf is at least one. This elementary observation does not establish Freiman's lower bound of three.