Supremum of Suprema

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

Let $T$ be a subset of the power set of $S$.

Suppose that for each $Q \in T$, $Q$ has a supremum in $S$.

Then $\left\{{\sup Q \mid Q \in T}\right\}$ has a supremum iff $\bigcup T$ does, and if they have them,

then $\sup \left\{{\sup Q \mid Q \in T}\right\} = \sup \bigcup T$.

Part One
Let $a \in S$.

Suppose that $a \ge \sup Q$ for each $Q \in T$.

If $x \in \bigcup T$, then for some $R \in T$, $x \in R$ by the definition of union.

Then by supposition, $a \ge \sup R$.

Thus $a \ge x$ by the definition of supremum.

Part Two
Let $a \in S$.

Suppose that $a \ge x$ for each $x \in \bigcup T$.

Let $R \in T$.

Then $R \subset \bigcup T$, so $a \ge x$ for each $x \in R$.

By the definition of supremum, $a \ge \sup R$.

We have thus shown that the set of upper bounds of $\{\sup Q \mid Q \in T\} $ equals the set of upper bounds of $\sup \bigcup T$, proving the theorem.