Boundedness of Metric Space by Open Ball

Theorem

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

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

Then $M'$ is bounded in $M$ if and only if:

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

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

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

Proof

Necessary Condition

Let $M' = \struct {Y, d_Y}$ be bounded in $M$ in the sense that:

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

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.

$\Box$

Sufficient Condition

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.

$\blacksquare$

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