Equivalence of Definitions of Bounded Metric Space

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Theorem

The following definitions of the concept of Bounded Metric Space are equivalent:

Definition 1

$M'$ is bounded (in $M$) if and only if:

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

That is, there exists an element of $A$ within a finite distance of all elements of $B$.

Definition 2

$M'$ is bounded if and only if:

$\exists K \in \R: \forall x, y \in M': d \left({x, y}\right) \le K$

That is, there exists a finite distance such that all pairs of elements of $B$ are within that distance.


Proof

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

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


Definition 1 implies Definition 2

Let $M'$ be bounded according to Definition 1:

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

Let $x, y \in Y$.

Then:

\(\displaystyle \left({x, y}\right)\) \(\le\) \(\displaystyle d \left({x, a}\right) + d \left({a, y}\right)\) Metric Space Axiom $M2$
\(\displaystyle \) \(=\) \(\displaystyle d \left({x, a}\right) + d \left({y, a}\right)\) Metric Space Axiom $M3$
\(\displaystyle \) \(\le\) \(\displaystyle K + K\) by hypothesis
\(\displaystyle \) \(=\) \(\displaystyle 2 K\) where $K$

$\blacksquare$

Thus:

$\left({x, y}\right) \le 2 K$


Thus, setting $r = 2 K$, $M'$ fulfils the conditions to be bounded according to Definition 2.

$\Box$

Definition 2 implies Definition 1

Let $M'$ be bounded according to Definition 2:

$\exists K \in \R: \forall x, y \in M': d \left({x, y}\right) \le K$

Let $a = y$.

Then:

$d \left({x, a}\right) \le K$

and so:

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

As $X \subseteq Y$ it follows by definition of subset that:

$\exists a \in Y \implies \exists a \in X$

and so:

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


Thus $M'$ fulfils the conditions to be bounded according to Definition 1.

$\blacksquare$