# Supremum of Set Equals Maximum of Supremums of Subsets It has been suggested that this page or section be merged into Maximum of Supremums of Subsets Equals Supremum of Set(to be removed). (Discuss)

## 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(to be removed).

$\blacksquare$