Completion Theorem (Metric Space)/Lemma 3

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:
 * $\tilde M = \left({\tilde A, \tilde d}\right)$ is a completion of $M$.

Proof
We are to show that:


 * $(1): \quad \tilde M$ is a complete metric space
 * $(2): \quad A \subseteq \tilde A$
 * $(3): \quad A$ is dense in $\tilde M$
 * $(4): \quad \forall x, y \in A : \tilde d \left({x, y}\right) = d \left({x, y}\right)$

For $x \in A$, let $\hat x = \left({x, x, x, \ldots}\right)$ be the constant sequence with value $x$.

Let $\phi: A \to \tilde A: x = \left[{\hat x}\right]$.

We first demonstrate that $(2)$ holds, by showing that $A \subseteq \tilde A$.

Thus:
 * $A \subseteq \tilde A$

Henceforth we identify $A$ with its isomorphic copy in $\tilde A$ when it is convenient.

Now we demonstrate that $(4)$ holds, by showing that $\phi$ is an injection from $A$ into $\tilde A$.

For any $x, y \in A$:

That is:
 * $\forall x, y \in A : \tilde d \left({x, y}\right) = d \left({x, y}\right)$

Now we demonstrate that $(3)$ holds, by showing that $A$ is dense in $\tilde A$.

Recall that the closure of $A$ is the union of $A$ and the limit points of $A$.

Let $\left[{x_n}\right] \in \tilde A$ and $\epsilon > 0$ be arbitrary.

If we can find $x \in A$ such that $\tilde d \left({\left[{\hat x}\right], \left[{x_n}\right]}\right) < \epsilon$ then we have shown that $A$ is dense in $\tilde A$.

Since $\left\langle{x_n}\right\rangle$ is Cauchy, there exists $N \in \N$ such that:
 * $\forall m, n \ge N: d \left({x_m, x_n}\right) < \epsilon$

Then we have:

and therefore $A$ is dense in $\tilde A$.

Finally we demonstrate that $(1)$ holds, by showing that $\left({\tilde A, \tilde d}\right)$ is complete.

By the completeness criterion it is sufficient to show that every Cauchy sequence in $\phi \left({A}\right)$ converges in $\tilde A$.

Let $\left\langle{\hat w_n}\right\rangle$ be a Cauchy sequence in $\phi \left({A}\right)$, so each $\hat w_n$ has the form $\left\langle{w_n, w_n, w_n, \ldots}\right\rangle$.

Since $\phi$ is an isometry:
 * $\forall m, n \in \N: \tilde d \left({\hat w_n, \hat w_m}\right) = d \left({w_n, w_m}\right)$

Therefore, $\left\langle{w_1, w_2, w_3,\ldots}\right\rangle$ is Cauchy in $A$.

Let $W = \left[{\left\langle{w_1, w_2, w_3, \ldots}\right\rangle}\right] \in \tilde A$.

We claim that $\left\langle{\hat w_n}\right\rangle$ converges to $W$ in $\tilde A$.

Let $\epsilon > 0$ be arbitrary.

Since $\left\langle{w_1, w_2, w_3, \ldots}\right\rangle$ is Cauchy in $A$, there exists $N \in \N$ such that for all $m, n \ge N$, we have $d \left({w_n, w_m}\right) < \epsilon$.

Thus for all $n > N$:


 * $\displaystyle \tilde d \left({w_n, W}\right) = \lim_{n \mathop \to \infty} d \left({w_n, W}\right) < \epsilon$

Therefore, $\left\langle{\hat w_n}\right\rangle \to W$ as $N \to \infty$, and $\tilde A$ is complete.