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 $\sqbrk {x_n}$.

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 Metric Space Axiom $\text M 4$

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 Metric Space Axiom $\text M 4$ holds for $\tilde d$.


Proof of Metric Space Axiom $\text M 1$

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

$\Box$


Proof of Metric Space Axiom $\text M 3$

We have that:

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

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

$\Box$


Proof of Metric Space Axiom $\text M 2$

We have that:

\(\ds \map {\tilde d} {\sqbrk {x_n}, \sqbrk {z_n} }\) \(=\) \(\ds \lim_{n \mathop \to \infty} \map d {x_n, z_n}\)
\(\ds \) \(\le\) \(\ds \lim_{n \mathop \to \infty} \set{\map d {x_n, y_n} + \map d {y_n, z_n} }\) $d$ 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
\(\ds \) \(=\) \(\ds \map {\tilde d} {\sqbrk {x_n}, \sqbrk {y_n} } + \map {\tilde d} {\sqbrk {y_n}, \sqbrk {z_n} }\)

So Metric Space Axiom $\text M 2$ holds for $\tilde d$.

$\Box$


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

$\blacksquare$