Maximum of Supremums of Subsets Equals Supremum of Set

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Theorem

Let $S$ be a non-empty real set.

Let $S$ have a supremum.

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

Let $\bigcup S_i = S$.


Then every $S_i$ has a supremum and:

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


Proof

By Supremum of Set of Real Numbers is at least Supremum of Subset, $\sup S_i$ exists for every $i$ in $\left\{{1, 2, \ldots, n}\right\}$.

Also, $\sup S \ge \sup S_i$ for every $i$ in $\left\{{1, 2, \ldots, n}\right\}$.

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

By Supremum of Subset of Union Equals Supremum of Union, $\sup S = \sup S_j$ for a $j$ in $\left\{{1, 2, \ldots, n}\right\}$.

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

$\blacksquare$


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