Upper Closure is Smallest Containing Upper Section

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

Let $T \subseteq S$.

Let $U = T^\succeq$ be the upper closure of $T$.

Then $U$ is the smallest upper set containing $T$ as a subset.

Proof
Follows from Upper Closure is Closure Operator and Set Closure is Smallest Closed Set/Closure Operator.

Also see

 * Lower Closure is Smallest Containing Lower Set