Complete and Totally Bounded Metric Space is Sequentially Compact/Proof 4

Proof
We use the following lemma.

Lemma
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $A$.

In the following, we will construct recursively nested subsequences:
 * $\sequence {x ^{\paren 0} _n} _{n \mathop \in \N}\supseteq \sequence {x ^{\paren 1} _n} _{n \mathop \in \N}\supseteq \sequence {x ^{\paren 2} _n} _{n \mathop \in \N}\cdots$

Let:
 * $\sequence {x ^{\paren 0} _n} _{n \mathop \in \N} := \sequence {x_n}_{n \mathop \in \N}$

For $m\in\Z _{>0}$, we apply the above lemma to $\sequence {x ^{\paren {m-1}} _n} _{n \mathop \in \N}$.

Then:
 * $\exists c_m\in A$

and:
 * $\exists \sequence {x ^{\paren m} _n} _{n \mathop \in \N}\subseteq\sequence {x ^{\paren {m-1}} _n} _{n \mathop \in \N}$ a subsequence such that:
 * $\forall n\in\N : \map d {x ^{\paren m} _n, c_m} < \dfrac 1 m$

Now, let:
 * $y _n := x ^{\paren n} _n$

for all $n\in\N$.

By construction:
 * $\sequence {y _n} _{n \mathop \in \N}$ is a subsequence of $\sequence {x_n}_{n \mathop \in \N}$
 * $\forall m\in\Z _{>0}$: $\sequence {y _n} _{n \ge m}$ is a subsequence of $\sequence {x ^{\paren m} _n} _{n \mathop \in \N}$

Finally, we shall show $\sequence {y _n} _{n \mathop \in \N}$ is convergent.

As $\struct {A, d}$ is complete, it suffices to show it is a Cauchy sequence.

Let $\epsilon \in \R_{>0}$ be arbitrary.

Let $N\in\N$ so large that:
 * $\dfrac 2 N < \epsilon$

Then, by (M2) of metric axiom:
 * $\forall m,n\geq N : \map d {y_m, y_n} \le \map d {y_m, c_N} + \map d {y_n, c_N} < \dfrac 1 N + \dfrac 1 N < \epsilon$