Element in Bounded Metric Space has Bound

Theorem

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

Let $M' = \struct {Y, d_Y}$ be a subspace of $M$.

Let $M'$ be bounded in $M$.

Then:

$\forall a' \in X: \exists K' \in \R: \forall x \in Y: \map d {x, a'} \le K'$

That is, if there is one element of $X$ which satisfies the condition for $Y$ to be bounded in $M$, they all do.

Proof

Let $a \in X$ such that $\exists K \in \R: \forall x \in Y: \map d {x, a} \le K$.

Let $a' \in X$.

\(\ds \map d {x, a'}\)

\(\le\)

\(\ds \map d {x, a} + \map d {a, a'}\)

Metric Space Axiom $(\text M 2)$: Triangle Inequality

\(\ds \)

\(\le\)

\(\ds K + \map d {a, a'}\)

by hypothesis

\(\ds \)

\(=\)

\(\ds K'\)

where $K' = K + \map d {a, a'}$

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.2$: Examples: Definitions $2.2.12$