Supremum of Union of Bounded Above Sets of Real Numbers
Theorem
Let $A$ and $B$ be sets of real numbers.
Let $A$ and $B$ both be bounded above.
Then:
- $\map \sup {A \cup B} = \max \set {\sup A, \sup B}$
where $\sup$ denotes the supremum.
Proof
Let $A$ and $B$ both be bounded above.
By the Continuum Property, $A$ and $B$ both admit a supremum.
Let $x \in A \cup B$.
Then either $x \le \sup A$ or $x \le \sup B$ by definition of supremum.
Hence:
- $x \le \max \set {\sup A, \sup B}$
and so $\max \set {\sup A, \sup B}$ is certainly an upper bound of $A \cup B$.
It remains to be shown that $\max \set {\sup A, \sup B}$ is the smallest upper bound.
Aiming for a contradiction, suppose there exists $m \in \R$ such that:
- $m < \max \set {\sup A, \sup B}$
and:
- $\forall x \in A \cup B: x \le m$
Without loss of generality, let $\sup A \ge \sup B$.
Then:
- $\max \set {\sup A, \sup B} = \sup A$
and so:
- $m < \sup A$
But then by definition of supremum:
- $\exists a \in A: a > m$
and so $m$ is not an upper bound of $A \cup B$.
This contradicts our assumption that $m$ is a supremum of $A \cup B$.
It follows by Proof by Contradiction that $\max \set {\sup A, \sup B}$ is the supremum of $A \cup B$.
The same argument shows, mutatis mutandis, that if $\sup A \le \sup B$, $\max \set {\sup A, \sup B}$ is the supremum of $A \cup B$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 2$