Upper and Lower Closures are Convex
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Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $a \in S$.
Then $a^\succeq$, $a^\succ$, $a^\preceq$, and $a^\prec$ 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 $C = a^\succeq$ or $C = a^\succ$.
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 $y \in a^\succ \subseteq C$.
Thus $C$ is convex.
$\blacksquare$