Ray is Convex

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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.

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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.

$\blacksquare$

Also see

• Upper and Lower Closures are Convex