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 $\tilde d$. {{qed|lemma}} === Proof of $M3$ === We have that: {{begin-eqn}} {{eqn | l = \tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right)       | r = \lim_{n \mathop \to \infty} d \left({x_n, y_n}\right)       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} d \left({y_n, x_n}\right)       | c = $d$ is a Metric }} {{eqn | r = \tilde d \left({\left[{y_n}\right], \left[{x_n}\right]}\right)       | c = }} {{end-eqn}} So axiom $M3$ holds for $\tilde d$. {{qed|lemma}} === Proof of $M2$ === We have that: {{begin-eqn}} {{eqn | l = \tilde d \left({\left[{x_n}\right], \left[{z_n}\right]}\right)       | r = \lim_{n \mathop \to \infty} d \left({x_n, z_n}\right)       | c = }} {{eqn | o = \le       | r = \lim_{n \mathop \to \infty} \left\{ {d \left({x_n, y_n}\right) + d \left({y_n, z_n}\right)}\right\}       | c = $d$ is a metric, and using elementary properties of limits (Reference?) }} {{eqn | r = \lim_{n \mathop \to \infty} d\left({x_n, y_n}\right) +  \lim_{n \to \infty} d \left({y_n, z_n}\right)       | c = Sum Rule for Real Sequences }} {{eqn | r = \tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) + \tilde d \left({\left[{y_n}\right], \left[{z_n}\right]}\right) }} {{end-eqn}} So axiom $M2$ holds for $\tilde d$. {{qed|lemma}} Thus $\tilde d$ satisfies all the metric space axioms and so is a metric.