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We show that, for a positive integer $r$, every minimal 1-saturating set in${\rm PG}(r-1,2)$ of size at least ${11/36} 2^r+3$ is either a complete cap orcan be obtained from a complete cap $S$ by fixing some $s\in S$ and replacingevery point $s'\in S\setminus\{s\}$ by the third point on the line through $s$and $s'$. Stated algebraically: if $G$ is an elementary abelian 2-group and aset $A\subseteq G\setminus\{0\}$ with $|A|>{11/36} |G|+3$ satisfies $A\cup2A=G$ and is minimal subject to this condition, then either $A$ is a maximalsum-free set, or there are a maximal sum-free set $S\subseteq G$ and an element$s\in S$ such that $A=\{s\}\cup\big(s+(S\setminus\{s\})\big)$. Since,conversely, every set obtained in this way is a minimal 1-saturating set, andthe structure of large sum-free sets in an elementary 2-group is known, thisprovides a complete description of large minimal 1-saturating sets.  Our approach is based on characterizing those large sets $A$ in elementaryabelian 2-groups such that, for every proper subset $B$ of $A$, the sumset 2Bis a proper subset of 2A.

1-Saturating Sets, Caps and Round Sets in Binary Spaces