Complete and Totally Bounded Metric Space is Sequentially Compact/Proof 1

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Let $M$ be complete and totally bounded.

Then $M$ is sequentially compact.

Proof
Let $\left\langle{x_m}\right\rangle_{m \in \N}$ be an infinite sequence in $A$.

By the definition of a totally bounded metric space, we can use the axiom of countable choice to obtain a sequence $\left\langle{F_n}\right\rangle_{n \in \N}$ such that:
 * For all $n \in \N$, $F_n$ is a finite $2^{-n}$-net for $M$.

For all $n \in \N$ and $y \in F_n$, define:
 * $S_n \left({y}\right) = \left\{{m \in \N: d \left({x_m, y}\right) < 2^{-n}}\right\}$

It follows from the definition of a net that:
 * $(1): \quad \displaystyle \N = \bigcup_{y \mathop \in F_n} S_n \left({y}\right)$

For all $n \in \N$, define:
 * $G_n = \left\{{y \in F_n: S_n \left({y}\right)}\right.$ is infinite$\left.{}\right\}$

Since $F_n$ is finite by definition, it follows by $(1)$ that $G_n$ is non-empty.

For all $y \in G_n$, define:
 * $\displaystyle T_n \left({y}\right) = \left\{{z \in G_{n+1}: S_n \left({y}\right) \cap S_{n+1} \left({z}\right)}\right.$ is infinite$\left.{}\right\}$

By $(1)$, it follows from the distributivity of intersection over union that:
 * $\displaystyle S_n \left({y}\right) = \bigcup_{z \mathop \in F_{n+1}} \left({S_n \left({y}\right) \cap S_{n+1} \left({z}\right)}\right)$

Hence, by the definition of $G_n$, it follows that $T_n \left({y}\right)$ is non-empty.

Since the countable union of countable sets is countable, it follows that the disjoint union $\displaystyle \bigsqcup_{n \mathop \in \N} G_n$ is countable.

Using the axiom of countable choice, there exists a sequence $\left\langle{\phi_n: G_n \to G_{n+1}}\right\rangle_{n \in \N}$ of mappings such that:
 * $\forall n \in \N: \forall y \in G_n: \phi_n \left({y}\right) \in T_n \left({y}\right)$

Now, we use the principle of recursive definition to construct a strictly increasing sequence $\left\langle{m_k}\right\rangle_{k \in \N}$ in $\N$.

Let $y_0 \in G_0$, and let $m_0 \in S_0 \left({y_0}\right)$.

For all $k \in \N$, let $y_{k+1} = \left({\phi_k \circ \cdots \circ \phi_1 \circ \phi_0}\right) \left({y_0}\right)$, where $\circ$ denotes composition of mappings.

Let $m_{k+1} > m_k$ be the smallest natural number such that $m_{k+1} \in S_k \left({y_k}\right) \cap S_{k+1} \left({y_{k+1}}\right)$.

Such an $m_{k+1}$ exists by the well-ordering principle, and because $S_k \left({y_k}\right) \cap S_{k+1} \left({y_{k+1}}\right)$ is infinite by the definitions of $T_k \left({y_k}\right)$ and $\phi_k$.

Note that $m_k, m_{k+1} \in S_k \left({y_k}\right)$.

Let $\hat x_k = x_{m_k}$.

Let $i < j$, where $i, j \in \N$.

Then, by Sum of Infinite Geometric Progression:
 * $\displaystyle d \left({\hat x_i, \hat x_j}\right) \le \sum_{k \mathop = i}^{j-1} \left({d \left({\hat x_k, y_k}\right) + d \left({\hat x_{k+1}, y_k}\right)}\right) < \sum_{k \mathop = i}^{\infty} 2^{1-k} = 2^{2-i}$

Hence, by Power of Number less than One, the sequence $\left\langle{\hat x_k}\right\rangle_{k \in \N}$ is Cauchy.

By the assumption that $M$ is complete, the sequence $\left\langle{\hat x_k}\right\rangle$ converges in $M$.

Since $\left\langle{\hat x_k}\right\rangle$ is a convergent subsequence of $\left\langle{x_m}\right\rangle$, it follows that $M$ is sequentially compact by definition.