Ordering is Equivalent to Subset Relation/Lemma
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Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Then:
- $\forall a_1, a_2 \in S: \paren {a_1 \preceq a_2 \implies {a_1}^\preceq \subseteq {a_2}^\preceq}$
where ${a_1}^\preceq$ denotes the lower closure of $a_1$.
Proof
Let $a_1 \preceq a_2$.
Then by the definition of lower closure:
- $a_1 \in {a_2}^\preceq$
Let $a_3 \in {a_1}^\preceq$.
Then by definition:
- $a_3 \preceq a_1$
As an ordering is transitive, it follows that:
- $a_3 \preceq a_2$
and so:
- $a_3 \in {a_2}^\preceq$
This holds for all $a_3 \in {a_1}^\preceq$.
Thus by definition of subset:
- ${a_1}^\preceq \subseteq {a_2}^\preceq$
$\blacksquare$