Strict Upper Closure is Upper Section

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

Let $p \in S$.

Then $p^\succ$, the strict upper closure of $p$, is an upper set.

Proof
Let $u \in p^\succ$.

Let $s \in S$ with $u \preceq s$.

Then by the definition of strict upper closure:
 * $p \prec u$

Thus by Extended Transitivity:
 * $p \prec s$

So by the definition of strict upper closure:
 * $s \in p^\succ$

Since this holds for all such $u$ and $s$, $p^\succ$ is an upper set.

Also see

 * Upper Closure is Upper Set
 * Strict Lower Closure is Lower Set