Strict Lower Closure is Lower Section/Proof 1

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

Let $p \in S$.

Then ${\dot\downarrow} p$, the strict down-set of $p$, is a lower set.

 Let $l \in {\dot\downarrow}p$.

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

Then by the definition of strict down-set, $l \prec p$.

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

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

Since this holds for all such $l$ and $s$, ${\dot\downarrow} p$ is a lower set.

Also see

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