# Ordering is Equivalent to Subset Relation/Lemma

## 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:

${a_1}^\preceq \subseteq {a_2}^\preceq$

$\blacksquare$