Element in Bounded Metric Space has Bound
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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 $2.2.12$