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 $M$.

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

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

For all natural numbers $n \ge 1$, define the set:
 * $\SS_n = \set {F \subseteq A: \size F = n: \forall x, y \in F: x \ne y \implies \map d {x, y} \ge \epsilon}$

where $\size F$ denotes the cardinality of $F$.

We use the Principle of Mathematical Induction to prove that $\SS_n$ is non-empty.

It is vacuously true that any singleton $\set x \subseteq A$ is an element of $\SS_1$.

Since $A$ is non-empty by the definition of a metric space, it follows from Existence of Singleton Set that $\SS_1$ is non-empty.

Let $F \in \SS_n$.

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:
 * $\forall y \in F: \map d {x, y} \ge \epsilon$

Note that, by :
 * $x \notin F$

Consider the set:
 * $F' := F \cup \set x$.

Then:
 * $\size {F'} = n + 1$

and by :
 * $F' \in \SS_{n + 1}$

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

Therefore, using the axiom of countable choice, we can obtain an infinite sequence $\sequence {F_n}_{n \ge 1}$ such that:
 * $\forall n \in \N_{\ge 1}: F_n \in \SS_n$

From Countable Union of Countable Sets is Countable, there exists an injection:
 * $\ds \phi: \bigcup_{n \mathop \ge 1} F_n \to \N$

We now construct an infinite sequence $\sequence {x_n}_{n \ge 1}$ in $A$.

To do this, we use the Principle of Recursive Definition to define the sequence $\sequence {\tuple {x_1, x_2, \ldots, x_n} }_{n \ge 1}$ of ordered $n$-tuples.

Let $x_1 \in F_1$.

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

Define:
 * $D_n = \set {x \in F_{n + 1}: \forall y \in T_n: \map d {x, y} \ge \dfrac \epsilon 2}$

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

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

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

Let $y \in \map {C_n} x$.

Then it follows from:
 * the definition of $F_{n + 1}$
 * and :

that:
 * $\map d {x', y} \ge \map d {x, x'} - \map d {x, y} > \dfrac \epsilon 2$

Hence, $y \notin \map {C_n} {x'}$.

That is, the indexed family of sets:
 * $\sequence {\map {C_n} x}_{x \in F_{n + 1}}$

is pairwise disjoint.

Suppose that $D_n$ is empty.

That is:
 * $\forall x \in F_{n + 1}: \map {C_n} x$ 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:
 * $\ds \size {F_{n + 1} } \le \sum_{x \mathop \in F_{n + 1} } \size {\map {C_n} x} \le \size {T_n} < \size {F_{n + 1} }$

which is a contradiction.

From the well-ordering principle, we have that $\struct {\N, \le}$ 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:
 * $\forall m, n \in \N_{>0}: m \le n \implies \map d {x_{n + 1}, x_m} \ge \dfrac \epsilon 2$

Hence, by induction, it follows from that:
 * $\forall m, n \in \N_{>0}: m \ne n \implies \map d {x_m, x_n} \ge \dfrac \epsilon 2$

Therefore, the sequence $\sequence {x_n}$ has no Cauchy subsequence.

From Convergent Sequence in Metric Space is Cauchy Sequence, $\sequence {x_n}$ 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.