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

Proof
Let $\sequence {x_m}_{m \mathop \in \N}$ 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 $\sequence {F_n}_{n \mathop \in \N}$ 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:
 * $\map {S_n} y = \set {m \in \N: \map d {x_m, y} < 2^{-n} }$

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

For all $n \in \N$, define:
 * $G_n = \leftset {y \in F_n: \map {S_n} y}$ is infinite$\rightset{}$

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

For all $y \in G_n$, define:
 * $\ds \map {T_n} y = \leftset {z \in G_{n + 1}: \map {S_n} y \cap \map {S_{n + 1} } z}$ is infinite$\rightset{}$

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

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

From Countable Union of Countable Sets is Countable, it follows that the disjoint union $\ds \bigsqcup_{n \mathop \in \N} G_n$ is countable.

Using the axiom of countable choice, there exists a sequence $\sequence {\phi_n: G_n \to G_{n + 1} }_{n \mathop \in \N}$ of mappings such that:
 * $\forall n \in \N: \forall y \in G_n: \map {\phi_n} y \in \map {T_n} y$

Now, we use the Principle of Recursive Definition to construct a strictly increasing sequence $\sequence {m_k}_{k \mathop \in \N}$ in $\N$.

Let $y_0 \in G_0$.

Let $m_0 \in \map {S_0} {y_0}$.

For all $k \in \N$, let:
 * $y_{k + 1} = \map {\paren {\phi_k \circ \cdots \circ \phi_1 \circ \phi_0} } {y_0}$

where $\circ$ denotes composition of mappings.

Let $m_{k + 1} > m_k$ be the smallest natural number such that:
 * $m_{k + 1} \in \map {S_k} {y_k} \cap \map {S_{k + 1} } {y_{k + 1} }$

Such an $m_{k + 1}$ exists by:
 * the well-ordering principle

and because:
 * $\map {S_k} {y_k} \cap \map {S_{k + 1} } {y_{k + 1} }$ is infinite by the definitions of $\map {T_k} {y_k}$ and $\phi_k$.

Note that:
 * $m_k, m_{k + 1} \in \map {S_k} {y_k}$

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

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

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

Hence, by Sequence of Powers of Number less than One, the sequence $\sequence {\hat x_k}_{k \mathop \in \N}$ is Cauchy.

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

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