Upper and Lower Closures are Convex

Theorem
Let $(S, \preceq)$ be an ordered set.

Let $a \in S$.

Then ${\uparrow} a$, ${\bar\uparrow} a$, ${\downarrow} a$, and ${\bar\downarrow} a$ are convex in $S$.

Proof
The cases for upper and lower closures are dual, so we need only prove the case for upper closures.

Suppose, then, that $C = {\uparrow} a$ or $C = {\bar\uparrow} a$.

Suppose that $x, y, z \in S$, $x \prec y \prec z$, and $x, z \in C$.

Then $a \preceq x \prec y$, so $a \prec y$ by Extended Transitivity.

Therefore $y \in {\uparrow}a \subseteq C$.

Thus $C$ is convex.