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.

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对 3k − 4 定理的单组改进

$3k-4$ 定理是一个经典的结果，它断言如果 $A,\,B\subseteq \mathbb Z$ 是具有 \begin{equation}\label{hyp}|A+B|=| 的有限非空子集A|+|B|+r\leq |A|+|B|+\min\{|A|,\,|B|\}-3-\delta,\end{equation} 其中 $\delta=1 $如果$A$和$B$相互平移，否则$\delta=0$，则存在等差级数$P_A$和$P_B$的公差使得$A\subseteq P_A$, $B \subseteq P_B$、$|B|\leq |P_B|+r+1$ 和 $|P_A|\leq |A|+r+1$。这是弗里曼定理中为数不多的已知级数大小的确切界限的情况之一。从某种意义上说，上述假设是最好的可能，即存在仅具有基数的和集 $A+B$ 的例子，但 $A$ 和 $B$ 不能同时包含在短长度的算术级数中。在本文中，我们表明，上述假设可以被显着削弱，并且仍然对 $A$ 和 $B$ 集合之一产生相同的结论。具体来说，如果 $|B|\geq 3$，$s\geq 1$ 是唯一整数，其中 $$(s-1)s\left(\frac{|B|}{2}-1\right)+ s-1<|A|\leq s(s+1)\left(\frac{|B|}{2}-1\right)+s,$$ 和 \begin{equation}\label{hyp2} | A+B|=|A|+|B|+r< (\frac{|A|}{s}+\frac{|B|}{2}-1)(s+1),\end{方程然后我们证明存在一个等差数列 $P_B\subseteq \mathbb Z$ 和 $B\subseteq P_B$ 和 $|P_B|\leq |B|+r+1$。上述假设最有可能得出这样的结论（无需对 $A$ 进行额外假设）。|A|\leq s(s+1)\left(\frac{|B|}{2}-1\right)+s,$$ 和 \begin{equation}\label{hyp2} |A+B| =|A|+|B|+r< (\frac{|A|}{s}+\frac{|B|}{2}-1)(s+1),\end{equation} 然后我们证明有一个等差数列 $P_B\subseteq \mathbb Z$ 与 $B\subseteq P_B$ 和 $|P_B|\leq |B|+r+1$。上述假设最有可能得出这样的结论（无需对 $A$ 进行额外假设）。|A|\leq s(s+1)\left(\frac{|B|}{2}-1\right)+s,$$ 和 \begin{equation}\label{hyp2} |A+B| =|A|+|B|+r< (\frac{|A|}{s}+\frac{|B|}{2}-1)(s+1),\end{equation} 然后我们证明有一个等差数列 $P_B\subseteq \mathbb Z$ 与 $B\subseteq P_B$ 和 $|P_B|\leq |B|+r+1$。上述假设最有可能得出这样的结论（无需对 $A$ 进行额外假设）。