Erdős Problem #656

Erdős Problem 656: a shifted restricted sumset

Statement.

If $A\subseteq\mathbb N$ has positive upper density, must there exist an infinite $B\subseteq A$ and an integer $t$ such that \[ \{b+b':b,b'\in B,\ b\ne b'\}+t\subseteq A? \]

Claim and attribution.

Yes. We reconstruct the passage from the established dynamical recurrence theorem to the infinite sumset, retaining its external inputs explicitly. The source is Kra, Moreira, Richter, and Robertson, A proof of Erdős's $B+B+t$ conjecture, https://arxiv.org/abs/2206.12377, published in Communications of the AMS $2024$. The relevant inputs are Proposition 2.3 and Theorem 1.4(i) in the cited version; the extraction below is its Theorem 2.2. See also https://www.erdosproblems.com/656.

The two external inputs.

Positive upper density implies positive upper Banach density: choose intervals $[1,N_j]$, with $N_j\to\infty$, on which the density converges to a positive limit. These intervals form a Følner sequence, meaning that their relative symmetric difference with any fixed translate tends to zero.

The correspondence principle in Proposition 2.3 provides an ergodic measure-preserving homeomorphism $T$ of a compact metric space $X$, a point $a$, a clopen set $E$ of positive measure, and a Følner sequence along which $a$ is generic, with the exact identity \[ A=\{n\in\mathbb N:T^na\in E\}. \] This particular ergodic correspondence principle is an external input; genericity along the required sequence is not inferred merely from compactness.

Theorem 1.4(i) then supplies $x_1,x_2\in X$, $t\in\mathbb N$, and strictly increasing integers $n_j$ such that \[ x_1\in E,\qquad T^tx_2\in E,\qquad (T^{n_j}a,T^{n_j}x_1)\longrightarrow(x_1,x_2). \tag{1} \] This recurrence theorem contains the substantial ergodic part of the known solution. It is used here without claiming a new proof of that part.

Extracting infinitely many compatible integers.

Set $U=E$ and $V=T^{-t}E$, both open. Then $x_1\in U$, $x_2\in V$. Pass to a tail of the sequence in $1$ so that $T^{n_j}a\in U$ for all its terms. We choose $b_1<b_2<\cdots$ from this sequence, maintaining \[ T^{b_i}x_1\in V\quad\hbox{for every chosen }i, \qquad T^{b_i+b_j}a\in V\quad(i<j). \tag{2} \] For the first choice, $1$ gives a sufficiently late $b_1$ with $T^{b_1}x_1\in V$. Suppose $b_1,\ldots,b_r$ have been chosen. The open set \[ W=U\cap\bigcap_{i=1}^r T^{-b_i}V \] contains $x_1$, by the induction hypothesis. The convergence in $1$ lets us choose $b_{r+1}>b_r$ with \[ T^{b_{r+1}}a\in W,\qquad T^{b_{r+1}}x_1\in V. \] This preserves all conditions in $2$. Every selected integer is in $A$, and for $i<j$, \[ T^{b_i+b_j+t}a\in E, \qquad\text{so}\qquad b_i+b_j+t\in A. \] Thus $B=\{b_1,b_2,\ldots\}$ is the required infinite set. The argument constructs one infinite set, rather than merely arbitrarily large finite sets with incompatible shifts.

Two elementary checks on the statement.

If $A$ has density one, no dynamical theorem is needed and one can take $t=0$. Given finitely many chosen $b_i\in A$, the finite intersection \[ A\cap\bigcap_i(A-b_i) \] still has density one, since its complement is contained in a finite union of sets of density zero. Choose the next, larger element from this intersection and continue. In contrast, the shift cannot always be omitted: if $A$ is the positive odd integers, the sum of two of its elements is even, so no infinite $B\subseteq A$ works with $t=0$. Taking an odd shift removes this parity obstruction.

Scope.

The full answer is proved relative to the two specified established dynamical inputs. The infinite extraction, the density-one case, and the need for a shift are proved explicitly.

Completion estimate.

COMPLETION: 100%