Element in Bounded Metric Space has Bound

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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 $K' = K + d \left({a, a'}\right)$

$\blacksquare$


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