# Supremum of Subset of Union Equals Supremum of Union

## 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 there is a $j$ in $\left\{{1, 2, \ldots, n}\right\}$ such that:

- $\sup S_j = \sup S$.

## Proof

If $S$ equals $S_j$ for a $j$ in $\left\{{1, 2, \ldots, n}\right\}$, 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 $\left\{{1, 2, \ldots, n}\right\}$.

There are two alternatives; either:

- $\sup S > \sup S_i$ for every $i$ in $\left\{{1, 2, \ldots, n}\right\}$

or:

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

Suppose that:

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

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

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:

\(\displaystyle x\) | \(>\) | \(\displaystyle \sup S - \epsilon\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sup S - \left(\sup S - \max \left({\sup S_1, \sup S_2, \ldots, \sup S_n}\right)\right)\) | definition of $\epsilon$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \max \left({\sup S_1, \sup S_2, \ldots, \sup S_n}\right)\) |

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

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

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

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

We just concluded that the alternative:

- $\sup S > \sup S_i$ for every $i$ in $\left\{{1, 2, \ldots, n}\right\}$

is not true.

Therefore, the other alternative:

- $\sup S = \sup S_j$ for a $j$ in $\left\{{1, 2, \ldots, n}\right\}$

is true.

$\blacksquare$