# Supremum of Set Equals Maximum of Suprema of Subsets

## Theorem

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

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 $S = \bigcup S_i$.

Then:

$S_i$ has a supremum for every $i$ in $\set {1, 2, \ldots, n}$
$S$ has a supremum

and, in either case:

$\sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$

## Proof

### Necessary Condition

Let:

$S$ have a supremum.

We need to show that:

$(1): \quad S_i$ has a supremum for every $i$ in $\set {1, 2, \ldots, n}$
$(2): \quad \sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$

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

This proves $(1)$.

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

Therefore:

$\sup S \ge \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$

In addition, by Supremum of Subset of Union Equals Supremum of Union:

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

Therefore:

$\sup S \le \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$

Accordingly:

$\sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$

This proves $(2)$.

$\Box$

### Sufficient Condition

Let:

$S_i$ have a supremum for every $i$ in $\set {1, 2, \ldots, n}$.

We need to show that:

$(3): \quad S$ has a supremum
$(4): \quad \sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$

Every $S_i$ has a supremum.

Therefore, $\max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ exists as a real number.

The number $\max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ is an upper bound for every $S_i$.

Therefore, $\max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ 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.

This proves $(3)$.

We have proved that $S$ has a supremum.

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

Therefore:

$\sup S \ge \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$

In addition, by Supremum of Subset of Union Equals Supremum of Union:

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

Therefore:

$\sup S \le \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$

Accordingly:

$\sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$

This proves $(4)$.

$\blacksquare$