# Sequentially Compact Metric Space is Totally Bounded/Proof 2

## Theorem

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

Let $M$ be sequentially compact.

Then $M$ is totally bounded.

## Proof

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

By definition, $M$ is totally bounded only if there exists a finite $\epsilon$-net for $M$.

We use a Proof by Contraposition.

To that end, suppose that there exists no finite $\epsilon$-net for $M$.

The aim is to construct an infinite sequence $\sequence {x_n} _{n \in \N}$ in $A$ that has no convergent subsequence.

Suppose that $x_0, x_1, \ldots, x_r \in A$ have been defined such that:

- $\forall m, n \in \set {0, 1, \ldots, r} : m \ne n \implies \map d {x_m, x_n} \ge \epsilon$

That is, any two distinct elements of $\set {x_0, x_1, \ldots, x_r}$ are at least $\epsilon$ apart.

By hypothesis, there exists no finite $\epsilon$-net for $M$.

Specifically, $\set {x_0, x_1, \ldots, x_r}$ is therefore not an $\epsilon$-net for $M$.

So, by definition of a $\epsilon$-net:

- $\ds A \nsubseteq \bigcup_{i \mathop = 0}^r \map {B_\epsilon} {x_i}$

where $\map {B_\epsilon} {x_i}$ denotes the open $\epsilon$-ball of $x_i$ in $M$.

Thus there must exist $x_{r + 1} \in A$ such that:

- $\ds x_{r + 1} \notin \bigcup_{i \mathop = 0}^r \map {B_\epsilon} {x_i}$

That is:

- $\exists x_{r + 1} \in A: \forall m, n \in \set {0, 1, \ldots, r, r + 1} : m \ne n \implies \map d {x_m, x_n} \ge \epsilon$

Thus, by induction, the infinite sequence $\sequence {x_n}$ in $A$ has been constructed such that:

- $\forall m, n \in \N: m \ne n \implies \map d {x_m, x_n} \ge \epsilon$

Thus $\sequence {x_n}$ has no Cauchy subsequence.

Since a convergent sequence is Cauchy, it has no convergent subsequence either.

Thus, by definition, $M$ is not sequentially compact.

From Rule of Transposition, if $M$ is sequentially compact, then there exists no finite $\epsilon$-net for $M$.

So if $M$ is sequentially compact, it is certainly not totally bounded.

$\blacksquare$

## Axiom of Countable Choice

This theorem depends on the Axiom of Countable Choice.

Although not as strong as the Axiom of Choice, the Axiom of Countable Choice is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.

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Work In Progressuse only axiom of countable choice, not axiom of dependent choiceYou can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by completing it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{WIP}}` from the code. |

## Also see

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $7.2$: Sequential compactness: Proposition $7.2.9$