Bounded Metric Space is not necessarily Totally Bounded

From ProofWiki
Jump to navigation Jump to search


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

Then it is not necessarily the case that $M$ is totally bounded.

Proof 1

Let $M = \struct {\R, d}$ be the real number line with the Euclidean metric.

Let $M' = \struct {\R, \delta}$ be the unity-bounded metric space on $M$ where $\delta$ is defined as:

$\delta = \dfrac d {1 + d}$

From Unity-Bounded Metric Space is Bounded, $M'$ is a bounded metric space.

From Unity-Bounded Metric Space on Real Number Line is not Totally Bounded, $M'$ is not a totally bounded metric space.


Proof 2

Let $d$ be a discrete metric on the open unit interval $\Bbb I := \openint 0 1 \subseteq \R$.

We have that for all $x \in \openint 0 1$ and for all $r \in \R_{> 1}$:

$\map {B_r} x = \openint 0 1$

where $\map {B_r} x$ denotes the open $r$-ball of $x$.

Thus $\struct {\Bbb I, d}$ is bounded.

Let $\epsilon \in \R_{>0}$ be a strictly positive real number such that $\epsilon < 1$.

Let $x \in \openint 0 1$ be arbitrary.

Then $\map {B_\epsilon} x$ contains $x$ alone.

So, there is no finite $\epsilon$-net for $\openint 0 1$.

By definition of totally bounded metric space, $\struct {\Bbb I, d}$ is not totally bounded.