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Conjecture: A set $X \subseteq \mathbb{R}^n$ is convex if and only if the following holds.

For any $x \in X$ and any vector $v \in \mathbb{R}^n$ such that $x+v \notin X$, it holds that for any scalar $a$ with $1 \leq a$ we have that $x+av \notin X$.

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Try to restate your condition in contrapositive way, and see what it says. – user53153 Jan 1 '13 at 22:26
Well the contrapositive basically says $\forall x,v,a (x \in X, v \in R^n, 1 \leq a, x+av \in X \rightarrow x+v \in X)$. I don't know how to manipulate it any further.... – goblin Jan 1 '13 at 22:49
Introduce a letter $y=x+av$ and write the vector $x+v$ in terms of $x$ and $y$. – user53153 Jan 1 '13 at 22:50
Draw a picture. – copper.hat Jan 1 '13 at 23:11
@copper.hat: can I borrow your $n$-dimensional graph paper? – robjohn Jan 1 '13 at 23:40
up vote 1 down vote accepted

Suppose X convex. Then if $x\in X$ and $x+v\notin X$, then $x+av\notin X$ for $a\geq1$ because X is convex and the contrary will imply that $x+v\in X$. On the other hand, suppose your condition. Take two points $x$ and $y$ in X. Then if $y+t(x-y)\notin X$ for some $t_{0}$ between $(0,1)$, we have $y+t(x-y)\notin X$ for all $t\geq t_{0}$. Then, we will must have $x\notin X$. Contradiction!

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