Completion Theorem (Metric Space)/Lemma 2

Lemma
Let $M = \left({A, d}\right)$ be a metric space.

Let $\mathcal C \left[{A}\right]$ denote the set of all Cauchy sequences in $A$.

Define the equivalence relation $\sim$ on $\mathcal C \left[{A}\right]$ by:


 * $\displaystyle \left\langle{x_n}\right\rangle \sim \left\langle{y_n}\right\rangle \iff \lim_{n \mathop \to \infty} d \left({x_n, y_n}\right) = 0$

Denote the equivalence class of $\left\langle{x_n}\right\rangle \in \mathcal C \left[{A}\right]$ by $\left[{x_n}\right]$.

Denote the set of equivalence classes under $\sim$ by $\tilde A$.

Define $\tilde d: \tilde A \to \R_{\ge 0}$ by:


 * $\displaystyle \tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) = \lim_{n \mathop \to \infty} d \left({x_n, y_n}\right)$

Then:
 * $\tilde d$ is a metric on $\tilde A$.

Proof
To prove $\tilde d$ is a metric, we verify that it satisfies the axioms $M1$, $M2$, $M3$ and $M4$.

Proof of $M4$
Let $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) = \infty$.

Then $\left\langle{x_n}\right\rangle$ and $\left\langle y_n \right\rangle$ cannot both be Cauchy.

So $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) < \infty$ for $\left[{x_n}\right], \left[{y_n}\right] \in \tilde A$.

By the definition of $\tilde d$, for any $\left[{x_n}\right], \left[{y_n}\right] \in \tilde A$, $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right)$ must be a limit point of $R_{\ge 0}$.

The closure of $\R_{\ge 0}$ is $\R_{\ge 0}$, so $\tilde d: \tilde A \times \tilde A \to \R_{\ge 0}$.

So axiom $M4$ holds for $\tilde d$.

Proof of $M1$
Let $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) = 0$, which means that:


 * $\displaystyle \lim_{n \mathop \to \infty} d\left({x_n, y_n}\right) = 0$

So by definition:
 * $\left\langle{x_n}\right\rangle \sim \left\langle{y_n}\right\rangle$

and:
 * $\left[{x_n}\right] = \left[{y_n}\right]$

As $d$ is a metric, we also find immediately:
 * $\tilde d \left({\left[{x_n}\right], \left[{x_n}\right]}\right) = 0$

So axiom $M1$ holds for $\tilde d$.

Proof of $M3$
We have that:

So axiom $M3$ holds for $\tilde d$.

Proof of $M2$
We have that:

So axiom $M2$ holds for $\tilde d$.

Thus $\tilde d$ satisfies all the metric space axioms and so is a metric.