Ray is Convex

Theorem
Let $\struct {S, \preceq}$ be an ordered set.

Let $I$ be a ray, either open or closed.

Then $I$ is convex in $S$.

Proof
The cases for upward-pointing and downward-pointing rays are equivalent.

, suppose that $U$ is an upward-pointing ray.

By the definition of a ray, there exists an $a \in S$ such that:
 * $I = a^\succ$

or;
 * $I = a^\succeq$

according to whether $U$ is open or closed.

Let $x, y, z \in S$ such that $x \prec y \prec z$ and $x, z \in I$.

Then:
 * $a \preceq x \prec y$

so:
 * $a \prec y$

Therefore:
 * $y \in a^\succ \subseteq I$

Thus $I$ is convex.

Also see

 * Upper and Lower Closures are Convex