Infimum of Subset of Bounded Below Set of Real Numbers

Theorem

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

Let $B$ be bounded below.

Then:

$\inf A \ge \inf B$

where $\inf$ denotes the infimum.

Proof

Let $B$ be bounded below.

By the Continuum Property, $B$ admits an infimum.

By Subset of Bounded Below Set is Bounded Below, $A$ is also bounded below.

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

Aiming for a contradiction, suppose $\inf A < \inf B$.

Then:

$\exists y \in A: y < \inf B$

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

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 $1.5: 3$