The continuous postage stamp problem

For a real set $A$ consider the semigroup $S(A)$, additively generated by $A$; that is, the set of all real numbers representable as a (finite) sum of elements of $A$. If $A \subset (0,1)$ is open and non-empty, then $S(A)$ is easily seen to contain all sufficiently large real numbers, and we let $G(A) := \sup \{u \in R \colon u \notin S(A) \}$. Thus, $G(A)$ is the smallest number with the property that any $u>G(A)$ is representable as indicated above. We show that if the measure of $A$ is large, then $G(A)$ is small; more precisely, writing for brevity \alpha := \mes A we have G(A) \le (1-\alpha) \lfloor 1/\alpha \rfloor \quad &\text{if $0 < \alpha \le 0.1$}, (1-\alpha+\alpha\{1/\alpha\})\lfloor 1/\alpha\rfloor \quad &\text{if $0.1 \le \alpha \le 0.5$}, 2(1-\alpha) \quad &\text{if $0.5 \le \alpha \le 1$}. Indeed, the first and the last of these three estimates are the best possible, attained for $A=(1-\alpha,1)$ and $A=(1-\alpha,1)\setminus\{2(1-\alpha)\}$, respectively; the second is close to the best possible and can be improved by $\alpha \{1/\alpha\} \lfloor 1/\alpha \rfloor \le \{1/\alpha\}$ at most. The problem studied is a continuous analogue of the linear Diophantine problem of Frobenius (in its extremal settings due to Erdos and Graham), also known as the "postage stamp problem" or the "coin exchange problem"

A Single Set Improvement to the 3k−43k-4 Theorem

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