# Ray is Convex

## Theorem

Let $\left({S, \preceq}\right)$ 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.

Without loss of generality, 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.

$\blacksquare$