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:
\(\ds x\) | \(>\) | \(\ds \sup S - \epsilon\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sup S - \paren {\sup S - \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n} }\) | definition of $\epsilon$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n}\) |
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.
$\blacksquare$