Upper Closure is Upper Section

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

Let $T$ be a subset of $S$.

Let $U$ be the upper closure of $T$.

Then $U$ is an upper set.

Proof
Let $a \in U$.

Let $b \in S$ with $a \preceq b$.

By the definition of upper closure, there is a $t \in T$ such that $t \preceq a$.

By transitivity, $t \preceq b$.

Thus, agin by the definition of upper closure, $b \in U$.

Since this holds for all such $a$ and $b$, $U$ is an upper set.

Also see

 * Lower Closure is Lower Set