Upper Closure is Smallest Containing Upper Section

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

Let $T \subseteq S$.

Let $U = {\uparrow} T$ be the upper closure of $T$.

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

Proof
By Upper Closure is Upper Set, $U$ is an upper set in $S$.

Let $V$ be an upper set in $S$ and suppose that $T \subseteq V$.

Let $u \in U$.

By the definition of upper closure, there is some $t \in T$ such that $u \in {\bar\uparrow}t$, where ${\bar\uparrow}t$ is the upper closure of $t$.

Then $t \preceq u$.

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

Since $V$ is an upper set, $u \in V$.

Since each element of $U$ is in $V$, $U \subseteq V$.

Also see

 * Lower Closure is Smallest Containing Lower Set