Upper Closure is Closure Operator

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

Let ${\uparrow} T$ be the upper closure of $T$ for each $T \subseteq S$.

Then $\uparrow$ is a closure operator.

Extensive
Let $T \subseteq S$.

By Upper Closure is Upper Set: ${\uparrow} T$ is an upper set.

Thus $T = {\uparrow} T$.

By Equality of Sets, ${\uparrow} T \subseteq T$.

Since this holds for all $T \subseteq S$, $\uparrow$ is extensive.

Increasing
Let $T \subseteq U \subseteq S$.

Let $x \in {\uparrow} T$.

Then by the definition of upper closure: for some $t \in T$, $t \preceq x$.

By the definition of subset, $t \in U$.

Thus by the definition of upper closure, $x \in {\uparrow} U$.

Since this holds for all such $x$, $T \subseteq U$.

Since this holds for all such $T$, $\uparrow$ is increasing.

Idempotent
Let $T \subseteq S$.

By Upper Closure is Upper Set, ${\uparrow}T$ is an upper set.

Thus by Equivalence of Upper Set Definitions, ${\uparrow}\left({\uparrow} T\right) \subseteq {\uparrow}T$.

Since $\uparrow$ is extensive (see above), ${\uparrow} T \subseteq {\uparrow}\left({\uparrow} T\right)$

Thus by Equality of Sets, ${\uparrow}\left({\uparrow} T\right) = {\uparrow}T$.

Since this holds for all $T$, $\uparrow$ is idempotent.

Also see

 * Lower Closure is Closure Operator