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.
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 $(4)$.
$\blacksquare$
Also see
- Supremum of Suprema (a more general result)