Completion Theorem (Metric Space)/Lemma 3
== 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:  :$\tilde M = \left({\tilde A, \tilde d}\right)$ is a completion of $M$.  == Proof == We are to show that: : $(1): \quad \tilde M$ is a complete metric space : $(2): \quad A \subseteq \tilde A$ : $(3): \quad A$ is dense in $\tilde M$ : $(4): \quad \forall x, y \in A : \tilde d \left({x, y}\right) = d \left({x, y}\right)$ For $x \in A$, let $\hat x = \left({x, x, x, \ldots}\right)$ be the constant sequence with value $x$. Let $\phi: A \to \tilde A: x = \left[{\hat x}\right]$. We first demonstrate that $(2)$ holds, by showing that $A \subseteq \tilde A$. {{begin-eqn}} {{eqn | l = \phi \left({x}\right)       | r = \phi \left({y}\right) }} {{eqn | ll= \implies       | l = \left[{\hat x}\right]       | r = \left[{\hat y}\right] }} {{eqn | ll= \implies       | l = \lim_{n \mathop \to \infty} d \left({x, y}\right)       | r = 0 }} {{eqn | ll= \implies       | l = d \left({x, y}\right)       | r = 0 }} {{eqn | ll= \implies       | l = x       | r = y }} {{end-eqn}} Thus: :$A \subseteq \tilde A$ {{qed|lemma}} Henceforth we identify $A$ with its isomorphic copy in $\tilde A$ when it is convenient. Now we demonstrate that $(4)$ holds, by showing that $\phi$ is an injection from $A$ into $\tilde A$. For any $x, y \in A$: {{begin-eqn}} {{eqn | l = \tilde d \left({\left[{\hat x}\right], \left[{\hat y}\right]}\right)       | r = \lim_{n \mathop \to \infty} d \left({x, y}\right) }} {{eqn | r = d \left({x, y}\right) }} {{end-eqn}} That is: :$\forall x, y \in A : \tilde d \left({x, y}\right) = d \left({x, y}\right)$ {{qed|lemma}} Now we demonstrate that $(3)$ holds, by showing that $A$ is dense in $\tilde A$. Recall that the closure of $A$ is the union of $A$ and the limit points of $A$. Let $\left[{x_n}\right] \in \tilde A$ and $\epsilon > 0$ be arbitrary. If we can find $x \in A$ such that $\tilde d \left({\left[{\hat x}\right], \left[{x_n}\right]}\right) < \epsilon$ then we have shown that $A$ is dense in $\tilde A$. Since $\left\langle{x_n}\right\rangle$ is Cauchy, there exists $N \in \N$ such that: :$\forall m, n \ge N: d \left({x_m, x_n}\right) < \epsilon$ Then we have: {{begin-eqn}} {{eqn | l = \tilde d \left({\left[{\hat x_N}\right], \left[{x_n}\right]}\right)       | r = \lim_{n \mathop \to \infty} d \left({x_N, x_n}\right) }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} and therefore $A$ is dense in $\tilde A$. {{qed|lemma}} Finally we demonstrate that $(1)$ holds, by showing that $\left({\tilde A, \tilde d}\right)$ is complete. By the completeness criterion it is sufficient to show that every Cauchy sequence in $\phi \left({A}\right)$ converges in $\tilde A$. Let $\left\langle{\hat w_n}\right\rangle$ be a Cauchy sequence in $\phi \left({A}\right)$, so each $\hat w_n$ has the form $\left\langle{w_n, w_n, w_n, \ldots}\right\rangle$. Since $\phi$ is an isometry: :$\forall m, n \in \N: \tilde d \left({\hat w_n, \hat w_m}\right) = d \left({w_n, w_m}\right)$ Therefore, $\left\langle{w_1, w_2, w_3,\ldots}\right\rangle$ is Cauchy in $A$. Let $W = \left[{\left\langle{w_1, w_2, w_3, \ldots}\right\rangle}\right] \in \tilde A$. We claim that $\left\langle{\hat w_n}\right\rangle$ converges to $W$ in $\tilde A$. Let $\epsilon > 0$ be arbitrary. Since $\left\langle{w_1, w_2, w_3, \ldots}\right\rangle$ is Cauchy in $A$, there exists $N \in \N$ such that for all $m, n \ge N$, we have $d \left({w_n, w_m}\right) < \epsilon$. Thus for all $n > N$: :$\displaystyle \tilde d \left({w_n, W}\right) = \lim_{n \mathop \to \infty} d \left({w_n, W}\right) < \epsilon$  Therefore, $\left\langle{\hat w_n}\right\rangle \to W$ as $N \to \infty$, and $\tilde A$ is complete.  {{ACC|Completeness Criterion (Metric Spaces)}}