Element in Bounded Metric Space has Bound

Theorem

Let $M = \left({X, d}\right)$ be a metric space.

Let $M' = \left({Y, d_Y}\right)$ be a subspace of $M$.

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

Then:

$\forall a' \in X: \exists K' \in \R: \forall x \in Y: d \left({x, a}\right) \le K'$

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 $\exists K \in \R: \forall x \in Y: d \left({x, a}\right) \le K$.

Let $a' \in X$.

 $\displaystyle \left({x, a'}\right)$ $\le$ $\displaystyle d \left({x, a}\right) + d \left({a, a'}\right)$ Metric Space Axiom $M2$ $\displaystyle$ $\le$ $\displaystyle K + d \left({a, a'}\right)$ by hypothesis $\displaystyle$ $=$ $\displaystyle K'$ where $K' = K + d \left({a, a'}\right)$

$\blacksquare$