# Supremum of Set Equals Maximum of Supremums of Subsets

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## Theorem

Let $\left\{S_i: i \in \left\{{1, 2, \ldots, n}\right\}\right\}$, $n \in \N_{>0}$, be a set of non-empty real sets.

Let $S_i$ have a supremum for every $i$ in $\left\{{1, 2, \ldots, n}\right\}$.

Let $S = \bigcup S_i$.

Then $S$ has a supremum and:

$\sup S = \max \left({\sup S_1, \sup S_2, \ldots, \sup S_n}\right)$.

## Proof

$S$ is a real set as $S$ is the union of real sets.

$S_1$ is non-empty.

Therefore, $S$ is non-empty as $S$ is a superset of $S_1$.

Every $S_i$ has a supremum.

Therefore, $\max \left({\sup S_1, \sup S_2, \ldots, \sup S_n}\right)$ exists as a real number.

The number $\max \left({\sup S_1, \sup S_2, \ldots, \sup S_n}\right)$ is an upper bound for every $S_i$.

Therefore, $\max \left({\sup S_1, \sup S_2, \ldots, \sup S_n}\right)$ is an upper bound for $S$ as $S = \bigcup S_i$.

By the Continuum Property $S$ has a supremum as $S$ is a non-empty real set with an upper bound.

The result follows by Maximum of Supremums of Subsets Equals Supremum of Set.

$\blacksquare$