Union of Non-Disjoint Bounded Subsets of Metric Space is Bounded

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $B$ and $C$ be bounded subsets of $M$ such that $B \cap C \ne \O$.

Let $\map \diam B$ and $\map \diam C$ denote the diameters of $B$ and $C$.


Then $B \cup C$ is a bounded subset of $M$ such that:

$\map \diam {B \cup C} \le \map \diam B + \map \diam C$


Proof

Aiming for a contradiction, suppose there exists $x, y \in B \cup C$ such that:

$\map d {x, y} > \map \diam B + \map \diam C$

$x$ and $y$ cannot both be in $B$ or $C$, otherwise either $\map d {x, y} \le \map \diam B$ or $\map d {x, y} \le \map \diam C$.

Without loss of generality, let $x \in B$ and $y \in C$.

Let $z \in B \cap C$.

This is possible because $B \cap C \ne \O$ by hypothesis.

Then:

\(\ds \map d {x, z}\) \(\le\) \(\ds \map \diam B\) Definition of Diameter of $B$
\(\ds \map d {z, y}\) \(\le\) \(\ds \map \diam C\) Definition of Diameter of $C$
\(\ds \leadsto \ \ \) \(\ds \map d {x, z} + \map d {z, y}\) \(\le\) \(\ds \map \diam B + \map \diam C\)
\(\ds \leadsto \ \ \) \(\ds \map d {x, y}\) \(\le\) \(\ds \map \diam B + \map \diam C\) Metric Space Axiom $(\text M 2)$: Triangle Inequality: $\map d {x, y} \le \map d {x, z} + \map d {z, y}$

This contradicts our supposition that $\map d {x, y} > \map \diam B + \map \diam C$.


Hence:

\(\ds \forall x, y \in B \cup C: \, \) \(\ds \map d {x, y}\) \(\le\) \(\ds \map \diam B + \map \diam C\)
\(\ds \leadsto \ \ \) \(\ds \sup \set {\map d {x, y}: x, y \in B \cup C}\) \(\le\) \(\ds \map \diam B + \map \diam C\) Definition of Supremum
\(\ds \leadsto \ \ \) \(\ds \map \diam {B \cup C}\) \(\le\) \(\ds \map \diam B + \map \diam C\) Definition of Diameter of Subset of Metric Space

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Union_of_Non-Disjoint_Bounded_Subsets_of_Metric_Space_is_Bounded&oldid=544234"