Completion Theorem (Metric Space)/Lemma 2

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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 a metric on $\tilde A$.


Proof

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


Proof of Metric Space Axiom $(\text M 4)$

Let $\map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim} = \infty$.

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

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

By definition of $\tilde d$, for any $\eqclass {x_n} \sim, \eqclass {y_n} \sim \in \tilde A$, $\map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim}$ 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 Metric Space Axiom $(\text M 4)$ holds for $\tilde d$.


Proof of Metric Space Axiom $(\text M 1)$

Let $\map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim} = 0$, 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:

$\eqclass {x_n} \sim = \eqclass {y_n} \sim$

As $d$ is a metric, we also find immediately:

$\map {\tilde d} {\eqclass {x_n} \sim, \eqclass {x_n} \sim} = 0$


So Metric Space Axiom $(\text M 1)$ holds for $\tilde d$.

$\Box$


Proof of Metric Space Axiom $(\text M 3)$

We have that:

\(\ds \map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim}\) \(=\) \(\ds \lim_{n \mathop \to \infty} \map d {x_n, y_n}\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \map d {y_n, x_n}\) $d$ is a Metric
\(\ds \) \(=\) \(\ds \map {\tilde d} {\eqclass {y_n} \sim, \eqclass {x_n} \sim}\)

So Metric Space Axiom $(\text M 3)$ holds for $\tilde d$.

$\Box$


Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality

We have that:

\(\ds \map {\tilde d} {\eqclass {x_n} \sim, \eqclass {z_n} \sim}\) \(=\) \(\ds \lim_{n \mathop \to \infty} \map d {x_n, z_n}\)
\(\ds \) \(\le\) \(\ds \lim_{n \mathop \to \infty} \set{\map d {x_n, y_n} + \map d {y_n, z_n} }\) $d$ is a metric, and using elementary properties of limits (Reference?)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \map d {x_n, y_n} + \lim_{n \mathop \to \infty} \map d {y_n, z_n}\) Sum Rule for Real Sequences
\(\ds \) \(=\) \(\ds \map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim} + \map {\tilde d} {\eqclass {y_n} \sim, \eqclass {z_n} \sim}\)

So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $\tilde d$.

$\Box$


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

$\blacksquare$

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