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 $\text M 2$
\(\ds \) \(\le\) \(\ds K + \map d {a, a'}\) by hypothesis
\(\ds \) \(=\) \(\ds K'\) where $K' = K + \map d {a, a'}$

$\blacksquare$


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