Completion Theorem (Metric Space)/Lemma 1

Lemma
Let $M = \struct {A, d}$ be a metric space.

Let $\CC \sqbrk A$ denote the set of all Cauchy sequences in $A$.

Define the equivalence relation $\sim$ on $\CC \sqbrk A$ by:


 * $\ds \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$

Denote the equivalence class of $\sequence {x_n} \in \CC \sqbrk A$ by $\eqclass {x_n} \sim$.

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

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


 * $\ds \map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim} = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$

Then:
 * $\tilde d$ is well-defined on $\tilde A$.

Proof
Let $\sequence {x_n}$, $\sequence {\hat x_n}$, $\sequence {y_n}$, $\sequence {\hat y_n} \in \CC \sqbrk A$ be such that:


 * $\sequence {x_n} \sim \sequence {\hat x_n}$


 * $\sequence {y_n} \sim \sequence {\hat y_n}$

We have:

By an identical argument, we can also show that:


 * $\map d {\hat x_n, \hat y_n} - \map d {x_n, y_n} \le \map d {x_n, \hat x_n} + \map d {\hat y_n, y_n}$

and therefore:


 * $\ds 0 \le \size {\map d {x_n, y_n} - \map d {\hat x_n, \hat y_n} } \le \map d {x_n, \hat x_n} + \map d {\hat y_n, y_n}$

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

$\ds \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \lim_{n \mathop \to \infty} \map d {\hat x_n, \hat y_n}$

But this precisely means that:
 * $\map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim} = \map {\tilde d} {\eqclass {\hat x_n} \sim, \eqclass {\hat y_n} \sim}$