Supremum of Subset of Union Equals Supremum of Union

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

Let $S$ have a supremum.

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

Let $\bigcup S_i = S$.

Then there exists a $j$ in $\set {1, 2, \ldots, n}$ such that:


 * $\sup S_j = \sup S$

Proof
If $S$ equals $S_j$ for a $j$ in $\set {1, 2, \ldots, n}$, it is trivially true that $\sup S = \sup S_j$.

Now assume that $S$ is unequal to $S_i$ for every $i$ in $\left\{{1, 2, \ldots, n}\right\}$.

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

There are two alternatives; either:


 * $\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$

or:


 * $\sup S = \sup S_j$ for at least one $j$ in $\set {1, 2, \ldots, n}$.

Suppose that:


 * $\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$

Let $\epsilon = \sup S - \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n}$.

We note that $\epsilon > 0$.

By Supremum of Subset of Real Numbers is Arbitrarily Close, $S$ has an element $x$ that satisfies:


 * $x > \sup S - \epsilon$

We have:

Therefore:
 * $x > \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n}$

This means that $x > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$.

However, $x$ must be an element of $S_j$ for some $j$ in $\set {1, 2, \ldots, n}$ as $x \in S$ and $S = \bigcup S_i$.

Accordingly, it is not true that $\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$.

We just concluded that the alternative:


 * $\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$

is not true.

Therefore, the other alternative:


 * $\sup S = \sup S_j$ for a $j$ in $\set {1, 2, \ldots, n}$

is true.