Strict Upper Closure is Upper Section

Theorem
Let $\struct {S, \preceq}$ be an ordered set.

Let $p \in S$.

Let $p^\succ$ denote the strict upper closure of $p$.

Then $p^\succ$ is an upper section.

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 section.

Also see

 * Upper Closure is Upper Section
 * Strict Lower Closure is Lower Section