Equivalence of Definitions of Bounded Metric Space

Proof
Let $M = \struct {X, d}$ be a metric space.

Let $M' = \struct {Y, d_Y}$ be a subspace of $M$.

Definition 1 implies Definition 2
Let $M'$ be bounded according to Definition 1:


 * $\exists a \in X, K \in \R: \forall x \in Y: \map d {x, a} \le K$

Let $x, y \in Y$.

Then:

Thus:
 * $\map d {x, y} \le 2 K$

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

Definition 2 implies Definition 1
Let $M'$ be bounded according to Definition 2:


 * $\exists K \in \R: \forall x, y \in M': \map d {x, y} \le K$

Let $a = y$.

Then:
 * $\map d {x, a} \le K$

and so:
 * $\exists a \in Y, K \in \R: \forall x \in Y: \map d {x, a} \le K$

As $X \subseteq Y$ it follows by definition of subset that:
 * $\exists a \in Y \implies \exists a \in X$

and so:
 * $\exists a \in X, K \in \R: \forall x \in Y: \map d {x, a} \le K$

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

Definition 1 implies Definition 3
Let $M'$ be bounded according to Definition 1:


 * $\exists a \in X, K \in \R: \forall x \in Y: \map d {x, a} \le K$

Although not specified, $K \le 0$ for the definition to make sense, as $\map d {x, a} \ge 0$.

Let $\map {B_{K + 1} } a$ be the open $K + 1$-ball of $x$.

By definition of open ball:
 * $\map {B_{K + 1} } a := \set {x \in M: \map d {x, a} < K + 1}$

Let $y \in S$.

Then by definition:
 * $\map d {y, a} \le K < K + 1$

and so:
 * $y \in \map {B_{K + 1} } a$

It follows by the definition of subset that:
 * $Y \subseteq \map {B_{K + 1} } a$

and so $Y$ can be fitted inside an open ball.

Thus $M'$ fulfils the conditions to be bounded according to Definition 3.

Definition 3 implies Definition 1
Let $M'$ be bounded according to Definition 3:
 * $\exists x \in A, \epsilon \in \R_{>0}: B \subseteq \map {B_\epsilon} x$

where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$.

Then by definition:
 * $\forall y \in Y: \map d {x, y} \le \epsilon$

and so the condition for boundedness in $M$ is fulfilled.

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

Definition 1 implies Definition 4
Suppose:
 * $\exists x \in M, \epsilon \in \R_{>0}: Y \subseteq \map {B_\epsilon} x$

where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$.

Then by definition:
 * $\forall y \in Y: \map d {x, y} \le \epsilon$

and so the condition for boundedness in $M$ is fulfilled.

This follows from Element in Bounded Metric Space has Bound.

Definition 4 implies Definition 1
This follows directly.