# Boundedness of Metric Space by Open Ball

## Contents

## Theorem

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

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

Then $M'$ is bounded in $M$ iff

- $\exists x \in M, \epsilon \in \R_{>0}: Y \subseteq B_\epsilon \left({x}\right)$

where $B_\epsilon \left({x}\right)$ is the open $\epsilon$-ball of $x$.

Simply put: a subspace is bounded iff it can be fitted inside an open ball.

## Proof

### Necessary Condition

Let $M' = \left({Y, d_Y}\right)$ be bounded in $M$ in the sense that:

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

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

Let $B_{K+1} \left({a}\right)$ be the open $K+1$-ball of $x$.

By definition of open ball:

- $B_{K+1} \left({a}\right) := \left\{{x \in M: d \left({x, a}\right) < K+1}\right\}$

Let $y \in S$.

Then by definition:

- $d \left({y, a}\right) \le K < K + 1$

and so:

- $y \in B_{K+1} \left({a}\right)$

It follows by the definition of subset that:

- $Y \subseteq B_{K+1} \left({a}\right)$

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

$\Box$

### Sufficient Condition

Suppose:

- $\exists x \in M, \epsilon \in \R_{>0}: Y \subseteq B_\epsilon \left({x}\right)$

where $B_\epsilon \left({x}\right)$ is the open $\epsilon$-ball of $x$.

Then by definition:

- $\forall y \in Y: d \left({x, y}\right) \le \epsilon$

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

$\blacksquare$

## Used as definition

Some sources use this to define a bounded set.

## Sources

- 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): $\text{III}$: Compactness