Supremum of Suprema

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

Let $\mathbb T$ be a collection of subsets of $S$.

Suppose all $T \in \mathbb T$ admit a supremum $\sup T$ in $S$.

Then:


 * $\sup \bigcup \mathbb T = \sup \left\{{\sup T: T \in \mathbb T}\right\}$

as soon as one of these two quantities exists.

Proof
Suppose that $s = \sup \bigcup \mathbb T \in S$.

By 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, $s = \sup \left\{{\sup T: T \in \mathbb T}\right\}$.

Suppose now that $r = \sup \left\{{\sup T: T \in \mathbb T}\right\} \in S$.

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, $r = \sup \bigcup \mathbb T$.