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.

Inflationary
Let $T \subseteq S$.

Let $t \in T$.

Then since $T \subseteq S$, $t \in S$ by the definition of subset.

Since $\preceq$ is reflexive, $t \preceq t$.

Thus by the definition of upper closure, $t \in {\uparrow} T$.

Since this holds for all $t \in T$, $T \subseteq {\uparrow} T$.

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

Order-Preserving
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 $x \in {\uparrow} T$, ${\uparrow} T \subseteq {\uparrow} U$.

Since this holds for all $T$ and $U$, $\uparrow$ is order-preserving.

Idempotent
Let $T \subseteq S$.

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

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

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

Also see

 * Lower Closure is Closure Operator