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