Strict Upper Closure is Upper Section

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

Let $p \in S$.

Then ${\dot\uparrow} p$, the strict up-set of $p$, is an upper set.

Proof
Let $u \in {\dot\uparrow}p$.

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

Then by the definition of strict up-set, $p \prec u$.

Thus by Extended Transitivity, $p \prec s$.

So by the definition of strict up-set, $s \in {\dot\uparrow} p$.

Since this holds for all such $u$ and $s$, ${\dot\uparrow} p$ is an upper set.

Also see

 * Upper Closure is Upper Set
 * Strict Down-Set is Lower Set