Sequentially Compact Metric Space is Totally Bounded/Proof 1

Theorem
Let $M = \left({A, d}\right)$ 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 contradiction.

That is, suppose that there exists no finite $\epsilon$-net for $M$.

The aim is to construct an infinite sequence $\left \langle {x_n} \right \rangle_{n \ge 1}$ in $A$ that has no convergent subsequence.

For all natural numbers $n \ge 1$, define the set:
 * $\mathcal S_n = \left\{{F \subseteq A: \left\vert{F}\right\vert = n, \, \forall x, y \in F: x \ne y \implies d \left({x, y}\right) \ge \epsilon}\right\}$

where $\left\vert{F}\right\vert$ denotes the cardinality of $F$.

We use the principle of mathematical induction to prove that $\mathcal S_n$ is non-empty.

It is vacuously true that any singleton $\left\{{x}\right\} \subseteq A$ is an element of $\mathcal S_1$.

Since $A$ is non-empty by the definition of a metric space, it follows from Existence of Singleton Set that $\mathcal S_1$ is non-empty.

Now, suppose that $F \in \mathcal S_n$.

By definition, $F$ is finite, and so $F$ is not an $\epsilon$-net for $M$, by hypothesis.

Hence, there exists an $x \in A$ such that:
 * $\forall y \in F: d \left({x, y}\right) \ge \epsilon$

Note that $x \notin F$, by axiom $\left({M1}\right)$ for a metric.

Consider the set $F' = F \cup \left\{{x}\right\}$.

Then $\left\vert{F'}\right\vert = n + 1$, and by axiom $\left({M3}\right)$ for a metric, it follows that $F' \in \mathcal S_{n+1}$.

Thus, we have proven that $\mathcal S_n$ is non-empty for all natural numbers $n \ge 1$.

Therefore, using the axiom of countable choice, we can obtain an infinite sequence $\left\langle{F_n}\right\rangle_{n \ge 1}$ such that $F_n \in \mathcal S_n$ for all natural numbers $n \ge 1$.

Since the countable union of countable sets is countable, there exists an injection $\displaystyle \phi: \bigcup_{n \mathop \ge 1} F_n \to \N$.

We now construct an infinite sequence $\left\langle{x_n}\right\rangle_{n \ge 1}$ in $A$.

To do this, we use the principle of recursive definition to define the sequence $\left\langle{\left({x_1, x_2, \ldots, x_n}\right)}\right\rangle_{n \ge 1}$ of ordered $n$-tuples.

Let $x_1 \in F_1$.

Suppose that $x_1, x_2, \ldots, x_n$ have been defined, and let:
 * $T_n = \left\{{x_1, x_2, \ldots, x_n}\right\}$

Define:
 * $\displaystyle D_n = \left\{{x \in F_{n+1}: \forall y \in T_n: d \left({x, y}\right) \ge \frac \epsilon 2}\right\}$

Using a proof by contradiction, we show that $D_n$ is non-empty.

Suppose that $D_n$ is empty, i.e.:
 * $\displaystyle \forall x \in F_{n+1}: \exists y \in T_n: d \left({x, y}\right) < \frac \epsilon 2$

Since $\left\vert{T_n}\right\vert < \left\vert{F_{n+1}}\right\vert$, there exists a $y \in T_n$ and distinct $x, x' \in F_{n+1}$ such that:
 * $\displaystyle d \left({x, y}\right) < \frac \epsilon 2, \, d \left({x', y}\right) < \frac \epsilon 2$

By axioms $\left({M2}\right)$ and $\left({M3}\right)$ for a metric, it follows that:
 * $\displaystyle d \left({x, x'}\right) \le d \left({x, y}\right) + d \left({x', y}\right) < \epsilon$

which contradicts the definition of $F_{n+1}$.

Hence, $D_n$ is non-empty.

From the well-ordering principle, we have that $\left({\N, \le}\right)$ is a well-ordered set.

Let $\le_{\phi}$ be the ordering induced by $\phi$.

Then $\le_{\phi}$ is a well-ordering.

We define $x_{n+1}$ as the (unique) $\le_{\phi}$-smallest element of $D_n$.

By construction, we have:
 * $\displaystyle \forall m, n \in \N_{>0}: m \le n \implies d \left({x_{n+1}, x_m}\right) \ge \frac \epsilon 2$

Hence, by induction, it follows from axiom $\left({M3}\right)$ for a metric that:
 * $\displaystyle \forall m, n \in \N_{>0}: m \ne n \implies d \left({x_m, x_n}\right) \ge \frac \epsilon 2$

Therefore, the sequence $\left\langle{x_n}\right\rangle$ has no Cauchy subsequence.

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

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

But this contradicts the original assumption that $M$ is sequentially compact.