Subset of Bounded Subset of Metric Space is Bounded
Jump to navigation
Jump to search
Theorem
Let $M = \struct {A, d}$ be a metric space.
Let $B$ be a bounded subset of $M$.
Let $\map \diam B$ denote the diameter of $B$.
Let $C \subseteq B$ be a subset of $B$.
Then $C$ is a bounded subset of $M$ such that:
- $\map \diam C \le \map \diam B$
Proof
\(\ds \forall x, y \in B: \, \) | \(\ds \map d {x, y}\) | \(\le\) | \(\ds \map \diam B\) | Definition of Diameter of Subset of Metric Space | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall x, y \in C: \, \) | \(\ds \map d {x, y}\) | \(\le\) | \(\ds \map \diam B\) | Definition of Subset | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \sup \set {x, y \in C: \map d {x, y} }\) | \(\le\) | \(\ds \map \diam B\) | Definition of Supremum | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \diam C\) | \(\le\) | \(\ds \map \diam B\) | Definition of Diameter of Subset of Metric Space |
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: Exercise $2.6: 4$