Let $\left\langle{x_m}\right\rangle_{m \mathop \in \N}$ 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 $\left\langle{F_n}\right\rangle_{n \in \N}$ 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. From 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: :$\forall n \in \N: \forall y \in G_n: \phi_n \left({y}\right) \in T_n \left({y}\right)$ {{explain|It is rather unclear what the connection with PoRD is here}} Now, we use the Principle of Recursive Definition to construct a strictly increasing sequence $\left\langle{m_k}\right\rangle_{k \mathop \in \N}$ in $\N$. Let $y_0 \in G_0$. 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 Sequence of Powers 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. {{qed}}