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

$\blacksquare$