Completion Theorem (Metric Space)/Lemma 1

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 well-defined on $\tilde A$.

Proof
Let $\left\langle{x_n}\right\rangle$, $\left\langle{\hat x_n}\right\rangle$, $\left\langle{y_n}\right\rangle$, $\left\langle{\hat y_n}\right\rangle \in \mathcal C \left[{A}\right]$ be such that:


 * $\left\langle{x_n}\right\rangle \sim \left\langle{\hat x_n}\right\rangle$


 * $\left\langle{y_n}\right\rangle \sim \left\langle{\hat y_n}\right\rangle$

We have:

By an identical argument, we can also show that:


 * $d \left({\hat x_n, \hat y_n}\right) - d \left({x_n, y_n}\right) \le d \left({x_n, \hat x_n}\right) + d \left({\hat y_n, y_n}\right)$

and therefore:


 * $\displaystyle 0 \le \left\vert{d \left({x_n, y_n}\right) - d \left({\hat x_n, \hat y_n}\right)}\right\vert \le d \left({x_n, \hat x_n}\right) + d \left({\hat y_n, y_n}\right)$

Passing to the limit $n \to \infty$ and using the Combination Theorem for Sequences we have shown that:

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

But this precisely means that:
 * $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) = \tilde d \left({\left[{\hat x_n}\right], \left[{\hat y_n}\right]}\right)$