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