Supremum of Subset of Bounded Above Set of Real Numbers

Theorem

Let $A$ and $B$ be sets of real numbers such that $A \subseteq B$.

Let $B$ be bounded above.

Then:

$\sup A \le \sup B$

where $\sup$ denotes the supremum.

Proof

Let $B$ be bounded above.

By the Continuum Property, $B$ admits a supremum.

By Subset of Bounded Above Set is Bounded Above, $A$ is also bounded above.

Hence also by the Continuum Property, $A$ also admits a supremum.

Aiming for a contradiction, suppose $\sup A > \sup B$.

Then:

$\exists y \in A: y > \sup B$

Thus by definition of supremum, $y \notin B$.

That is:

$A \nsubseteq B$

which contradicts our initial assumption that $A \subseteq B$.

Hence the result by Proof by Contradiction.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 1$