A Single Set Improvement to the $3k-4$ Theorem

TL;DR

This paper weakens the conditions of the classical $3k-4$ Theorem, showing that under broader circumstances, one can still find an arithmetic progression containing one of the sets with controlled size, advancing additive combinatorics.

Contribution

It introduces a significantly weaker hypothesis under which the containment of a set in an arithmetic progression can be guaranteed, extending the classical $3k-4$ Theorem.

Findings

Established a new bound for the size of the progression containing $B$.

Demonstrated the optimality of the hypothesis without additional assumptions.

Extended the applicability of the $3k-4$ Theorem to broader cases.

Abstract

The $3k-4$ Theorem is a classical result which asserts that if $A,\,B\subseteq \mathbb Z$ are finite, nonempty subsets with \begin{equation}\label{hyp}|A+B|=|A|+|B|+r\leq |A|+|B|+\min\{|A|,\,|B|\}-3-\delta,\end{equation} where $\delta=1$ if $A$ and $B$ are translates of each other, and otherwise $\delta=0$, then there are arithmetic progressions $P_A$ and $P_B$ of common difference such that $A\subseteq P_A$, $B\subseteq P_B$, $|B|\leq |P_B|+r+1$ and $|P_A|\leq |A|+r+1$. It is one of the few cases in Freiman's Theorem for which exact bounds on the sizes of the progressions are known. The hypothesis above is best possible in the sense that there are examples of sumsets $A+B$ having cardinality just one more, yet $A$ and $B$ cannot both be contained in short length arithmetic progressions. In this paper, we show that the hypothesis above can be significantly weakened and still yield the same conclusion for one of the sets $A$ and $B$. Specifically, if $|B|\geq 3$, $s\geq 1$ is the unique integer with $$(s-1)s\left(\frac{|B|}{2}-1\right)+s-1<|A|\leq s(s+1)\left(\frac{|B|}{2}-1\right)+s,$$ and \begin{equation}\label{hyp2} |A+B|=|A|+|B|+r< (\frac{|A|}{s}+\frac{|B|}{2}-1)(s+1),\end{equation} then we show there is an arithmetic progression $P_B\subseteq \mathbb Z$ with $B\subseteq P_B$ and $|P_B|\leq |B|+r+1$. The above hypothesis is best possible (without additional assumptions on $A$) for obtaining such a conclusion.