Ordering is Equivalent to Subset Relation/Lemma

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Then:
 * $\forall a_1, a_2 \in S: \left({a_1 \preceq a_2 \implies {\bar\downarrow}a_1 \subseteq {\bar\downarrow}a_2}\right)$

where ${\bar\downarrow}$ represents lower closure.

Proof
Let $a_1 \preceq a_2$.

Then by the definition of lower closure:


 * $a_1 \in {\bar\downarrow}a_2$.

Let $a_3 \in {\bar\downarrow}a_1$.

Then by definition, $a_3 \preceq a_1$.

As an ordering is transitive, it follows that


 * $a_3 \preceq a_2$ and so $a_3 \in {\bar\downarrow}a_2$.

As this holds for all $a_3 \in {\bar\downarrow}a_1$:


 * ${\bar\downarrow}a_1 \subseteq {\bar\downarrow}a_2$ by definition of subset.