Supremum of Lower Closure of Set

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

Let $T \subseteq S$.

Let $L = T^\preceq$ be the lower closure of $T$ in $S$.

Let $s \in S$

Then $s$ is the supremum of $T$ it is the supremum of $L$.

Proof
By Supremum and Infimum are Unique we need only show that $s$ is a supremum of $L$ it is a supremum of $T$.

Forward Implication
Let $s$ be a supremum of $T$.

$s$ is an upper bound of $L$:

Let $l \in L$.

Then by the definition of lower closure, there is a $t \in T$ such that $l \preceq t$.

By the definition of supremum, $s$ is an upper bound of $T$.

Thus $t \preceq s$.

Since $\preceq$ is transitive, $l \preceq s$.

Since this holds for all $l \in L$, $s$ is an upper bound of $L$.

Suppose that $u$ is an upper bound of $L$.

Then since $T \subseteq L$, $u$ is an upper bound of $T$.

Thus by the definition of supremum, $s \preceq u$.

So:
 * $s$ is an upper bound of $L$
 * $s$ precedes all upper bounds of $L$

Thus it follows that $s$ is the supremum of $L$.

Reverse Implication
Let $s$ be a supremum of $L$.

Then $s$ is an upper bound of $L$.

Since $T \subseteq L$, $s$ is an upper bound of $T$.

Let $u$ be any upper bound of $T$ and let $l \in L$.

Then by the definition of lower closure, there is a $t \in T$ such that $l \preceq t$.

Then since $u$ is an upper bound of $T$, $t \preceq u$.

Thus since $\preceq$ is transitive, $l \preceq u$.

Thus $u$ is an upper bound of $L$.

By the definition of supremum, $s \preceq u$.

Thus $s$ is an upper bound of $T$ which precedes every upper bound of $T$.

Therefore $s$ is the supremum of $T$.