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.

$\blacksquare$