Completion Theorem (Metric Space)/Lemma 2

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 $\sqbrk {x_n}$.

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

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


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

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

Proof
To prove $\tilde d$ is a metric, we verify that it satisfies the metric space axioms.

Proof of
Let $\map {\tilde d} {\sqbrk {x_n}, \sqbrk {y_n} } = \infty$.

Then $\sequence {x_n}$ and $\sequence {y_n}$ cannot both be Cauchy.

So $\map {\tilde d} {\sqbrk {x_n}, \sqbrk {y_n} } < \infty$ for $\sqbrk {x_n}, \sqbrk {y_n} \in \tilde A$.

By the definition of $\tilde d$, for any $\sqbrk {x_n}, \sqbrk {y_n} \in \tilde A$, $\map {\tilde d} {\sqbrk {x_n}, \sqbrk {y_n} }$ 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 holds for $\tilde d$.

Proof of
Let $\map {\tilde d} {\sqbrk {x_n}, \sqbrk {y_n} } = 0$, which means that:


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

So by definition:
 * $\sequence {x_n} \sim \sequence {y_n}$

and:
 * $\sqbrk {x_n} = \sqbrk {y_n}$

As $d$ is a metric, we also find immediately:
 * $\map {\tilde d} {\sqbrk {x_n}, \sqbrk {x_n} } = 0$

So holds for $\tilde d$.

Proof of
We have that:

So holds for $\tilde d$.

Proof of
We have that:

So holds for $\tilde d$.

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