Supremum of Suprema
Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Let $\mathbb T \subseteq \powerset S$, where $\powerset S$ is the power set of $S$.
Suppose all $T \in \mathbb T$ admit a supremum $\sup T$ in $S$.
Then:
- $\sup \bigcup \mathbb T = \sup {\set {\sup T: T \in \mathbb T} }$
if one of these two quantities exists (in $S$).
Proof
Suppose that $s = \sup \bigcup \mathbb T \in S$.
By Set is Subset of Union, $T \subseteq \bigcup \mathbb T$ for all $T \in \mathbb T$.
Hence by Supremum of Subset:
- $\forall T \in \mathbb T: \sup T \preceq s$
Suppose now that $a \in S$ satisfies:
- $\forall T \in \mathbb T: \sup T \preceq a$
Then by transitivity of $\preceq$:
- $\forall t \in T: t \preceq a$
Since this holds for any $T \in \mathbb T$, also:
- $\forall t \in \bigcup \mathbb T: t \preceq a$
Hence $s \preceq a$, by definition of supremum.
That is, $s = \sup {\set {\sup T: T \in \mathbb T} }$.
$\Box$
Suppose now that $r = \sup {\set {\sup T: T \in \mathbb T} } \in S$.
By definition of supremum, for all $T \in \mathbb T$ and $t \in T$:
- $t \preceq \sup T$
By transitivity of $\preceq$:
- $\forall T \in \mathbb T: \forall t \in T: t \preceq r$
Hence for all $t \in \bigcup \mathbb T$:
- $t \preceq r$
Suppose that $a \in S$ satisfies:
- $\forall t \in \bigcup \mathbb T: t \preceq a$
In particular, for any $T \in \mathbb T$, since $T \subseteq \bigcup \mathbb T$:
- $\sup T \preceq a$
and therefore by definition of supremum, also:
- $r \preceq a$
That is, $r = \sup \bigcup \mathbb T$.
$\blacksquare$