Let $\alpha$ be the supremum of all $\delta$ such that there is a sequence $\langle A_n\rangle_{n=1}^\infty$ of measurable subsets of $(0,1)$ with the property that each $A_n$ has measure at least $\delta$ and for all $n,m\in\mathbb{N}$, $A_n\cap A_m\cap A_{n+m}=\emp$. For $k\in\mathbb{N}$, let $\alpha_k$ be the corresponding supremum for finite sequences $\langle A_n\rangle_{n=1}^k$. We show that $\alpha=\displaystyle\lim_{k\to\infty}\alpha_k$ and find the exact value of $\alpha_k$ for $k\leq 41$. In the process of finding these exact values, we also determine exactly the number of maximal sum free subsets of $\nhat{k}$ for $k\leq 41$. We also investigate the size of sets $\langle A_x\rangle_{x\in S}$ with $A_x\cap A_y\cap A_{x+y}=\emp$ where $S$ is a subsemigroup of $\big((0,\infty),+\big)$.

The first author acknowledges support recieved from the National Science Foundation via Grant DMS-0554803