Sequentially Compact Metric Space is Totally Bounded/Proof 1

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.

Let $F \in \mathcal S_n$.

By definition, $F$ is finite.

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, by axiom $\left({M1}\right)$ for a metric:
 * $x \notin F$

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:
 * $F' \in \mathcal S_{n+1}$

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:
 * $\forall n \in \N_{\ge 1}: F_n \in \mathcal S_n$

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

Using a Proof by Contradiction, we show that $D_n$ is non-empty.

For all $x \in F_{n+1}$, define:
 * $C_n \left({x}\right) = \left\{{y \in T_n: d \left({x, y}\right) < \dfrac \epsilon 2}\right\}$

Let $x, x' \in F_{n+1}$ be distinct.

Let $y \in C_n \left({x}\right)$.

Then it follows from:
 * the definition of $F_{n+1}$
 * axioms $\left({M2}\right)$ and $\left({M3}\right)$ for a metric

that:
 * $d \left({x', y}\right) \ge d \left({x, x'}\right) - d \left({x, y}\right) > \dfrac \epsilon 2$

Hence, $y \notin C_n \left({x'}\right)$.

That is, the indexed family of sets:
 * $\left\langle{C_n \left({x}\right)}\right\rangle_{x \in F_{n+1}}$

is pairwise disjoint.

Suppose that $D_n$ is empty.

That is:
 * $\forall x \in F_{n+1}: C_n \left({x}\right)$ is non-empty

Then, from Cardinality is Additive Function, Finite Union of Sets in Additive Function, and Cardinality of Subset of Finite Set, we have:
 * $\displaystyle \left\vert{F_{n+1}}\right\vert \le \sum_{x \mathop \in F_{n+1}} \left\vert{C_n \left({x}\right)}\right\vert \le \left\vert{T_n}\right\vert < \left\vert{F_{n+1}}\right\vert$

which is a contradiction.

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

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

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

From Convergent Sequence is Cauchy Sequence, $\left\langle{x_n}\right\rangle$ has no convergent subsequence either.

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

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

Thus the assumption that there exists no finite $\epsilon$-net for $M$ was false.

Therefore, by definition, $M$ is totally bounded.

Hence the result.