# Supremum of Suprema

## Theorem

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

Let $\mathbb T \subseteq \mathcal P \left({S}\right)$, where $\mathcal P \left({S}\right)$ is the power set of $S$.

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

Then:

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

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

$\Box$

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

$\blacksquare$