# Equivalence of Definitions of Bounded Metric Space

## Theorem

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

### Definition 1

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

$\exists a \in A, K \in \R: \forall x \in B: \map {d} {x, a} \le K$

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

### Definition 2

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

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

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

### Definition 3

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

$\exists x \in A, \epsilon \in \R_{>0}: B \subseteq \map {B_\epsilon} x$

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

That is, $M'$ can be fitted inside an open ball.

### Definition 4

Let $a' \in A$.

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

$\exists K \in \R: \forall x \in B: \map {d} {x, a'} \le K$

## 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:

 $\ds \map d {x, y}$ $\le$ $\ds \map d {x, a} + \map d {a, y}$ Metric Space Axiom $(\text M 2)$: Triangle Inequality $\ds$ $=$ $\ds \map d {x, a} + \map d {y, a}$ Metric Space Axiom $(\text M 3)$ $\ds$ $\le$ $\ds K + K$ by hypothesis $\ds$ $=$ $\ds 2 K$ where $K$

Thus:

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

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:

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

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.

$\Box$

### 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.

$\Box$

### 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.

$\Box$

### 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.

$\Box$

### Definition 4 implies Definition 1

This follows directly.

$\blacksquare$