Complete and Totally Bounded Metric Space is Sequentially Compact/Proof 3

Proof
Let $M$ be both complete and totally bounded.

Let $\sequence {a_k}$ be any infinite sequence in $A$.

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

Let $x_1, \ldots, x_n \in X$ be a finite set of points such that:


 * $\ds A = \bigcup_{i \mathop = 1}^n \map {B_\epsilon} {x_i}$

where $\map {B_\epsilon} {x_i}$ represents the open $\epsilon$-ball of $x_i$.

This is known to exist as $M$ is totally bounded.

Then for every $k \in \N$, there is some $j_k \in \set {0, \dots, n}$ such that $\map d {a_k, x_{j_k} } \le \epsilon$.

For some $j \in \set {0, \dots, n}$, we must have $j_k = j$ for infinitely many $k$, and it follows by setting $x := x_{j_k}$.

Setting $x := x_{j_k}$, we see that:
 * $(1): \quad$ There is some $x \in X$ such that $\map d {a_k, x} \le \epsilon$ for infinitely many $k$.

Now let $\sequence {a_k}$ be any infinite sequence in $A$.

By $(1)$, there is some $x_1 \in X$ such that $\map d {a_k, x_1} \le \dfrac 1 2$ for infinitely many $k$.

Now we can apply $(1)$ to the subsequence of $\sequence {a_k}$ which consisting of those elements for which $\map d {a_k, x_1} \le \dfrac 1 2$.

Thus we can find $x_2 \in A$ such that infinitely many $k$ satisfy both $\map d {a_k, x_2} \le \dfrac 1 4$ and $\map d {a_k, x_1} \le \dfrac 1 2$.

Now we proceed inductively, to obtain a sequence $\sequence {x_m}$ with the property that there exist infinitely many $k$ such that, for $1 \le j \le m$:
 * $(2) \quad \map d {a_k, x_j} \le 2^{-j}$

Now define a subsequence $\sequence {a_{k_m} }$ inductively by letting $k_0$ be arbitrary, and choosing $k_{m + 1}$ minimal such that $k_{m + 1} > k_m$ and such that $(2)$ holds for $k = k_m$ and all $1 \le j \le m$.

Let $\epsilon > 0$, and choose $n$ sufficiently large that $\paren {\dfrac 1 2}^{n - 1} < \epsilon$.

Then:
 * $\map d {a_{k_r}, a_{k_s} } \le \map d {a_{k_r}, x_n} + \map d {a_{k_s}, x_n} \le 2 \cdot 2^{-n} < \epsilon$

whenever $r, s \ge n$.

So this subsequence is a Cauchy sequence and hence, because $M$ is complete by assumption, it is convergent.

Thus we see that $\sequence {a_k}$ has a convergent subsequence.

Hence, by definition, $M$ is sequentially compact.