Completeness Criterion (Metric Spaces)/Proof 2

Proof
Let $\left\langle{x_n}\right\rangle$ be a Cauchy sequence in $M$.

Since $A$ is dense, we can invoke the Axiom of Countable Choice to choose for each $n$ some $y_n \in A$ which is within $1 / n$ of $x_n$.

We will show that $\left\langle{y_n}\right\rangle$ is Cauchy.

Let $\epsilon > 0$.

Since $\left\langle{x_n}\right\rangle$ is Cauchy, there exists some $N'$ such that:
 * $\forall n, m \ge N': d \left({x_n, x_m}\right) < \frac 1 3 \epsilon$

Let $N \in \N$ be greater than $N'$ and $3 \epsilon^{-1}$.

Note that $n, m \ge N$ implies that both:


 * $(1): \quad n, m \ge N'$


 * $(2): \quad \dfrac 1 n, \dfrac 1 m \le \dfrac 1 N < \dfrac 1 3 \epsilon$

Thus for $n, m \ge N$, by the Triangle Inequality and the above:

Hence $\left\langle{y_n}\right\rangle$ is a Cauchy sequence in $A$.

By assumption, it converges to some limit $y \in S$.

Now, we verify that $\left\langle{x_n}\right\rangle$ converges to $y$ as well.

Let $\epsilon > 0$.

Since $\left\langle{y_n}\right\rangle$ converges, there exists some $N'$ such that:
 * $\forall n \ge N': d \left({y_n, y}\right) < \dfrac 1 2 \epsilon$

Let $N \in \N$ be greater than $N'$ and $2 \epsilon^{-1}$.

Note that $n \ge N$ implies that both:


 * $(1): \quad n \ge N'$


 * $(2): \quad \dfrac 1 n \le \dfrac 1 N < \dfrac 1 2 \epsilon$

Thus, for $n \ge N$, we have by the Triangle Inequality and above that:

Hence $\left\langle{x_n}\right\rangle$ converges to $y$.