Does the following condition characterise convexity of a set?

Question

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 $x+av \notin X$.

Answer 1

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!