Completion Theorem (Metric Space)/Lemma 3

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:
 * $\tilde M = \struct {\tilde A, \tilde d}$ 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 $\tilde M$
 * $(4): \quad \forall x, y \in A : \map {\tilde d} {x, y} = \map d {x, y}$

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

Let $\phi: A \to \tilde A: x = \sqbrk {\hat x}$.

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:
 * $\forall x, y \in A : \map {\tilde d} {x, y} = \map d {x, y}$

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 $\sqbrk {x_n} \in \tilde A$ and $\epsilon > 0$ be arbitrary.

If we can find $x \in A$ such that $\map {\tilde d} {\sqbrk {\hat x}, \sqbrk {x_n} } < \epsilon$ then we have shown that $A$ is dense in $\tilde A$.

Since $\sequence {x_n}$ is Cauchy, there exists $N \in \N$ such that:
 * $\forall m, n \ge N: \map d {x_m, x_n} < \epsilon$

Then we have:

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

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

By the completeness criterion it is sufficient to show that every Cauchy sequence in $\phi \sqbrk A$ converges in $\tilde A$.

Let $\sequence {\hat w_n}$ be a Cauchy sequence in $\phi \sqbrk A$, so each $\hat w_n$ has the form $\tuple {w_n, w_n, w_n, \ldots}$.

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

Therefore, $\tuple {w_1, w_2, w_3, \ldots}$ is Cauchy in $A$.

Let $W = \sqbrk {\tuple {w_1, w_2, w_3, \ldots} } \in \tilde A$.

We claim that $\sequence {\hat w_n}$ converges to $W$ in $\tilde A$.

Let $\epsilon > 0$ be arbitrary.

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

Thus for all $n > N$:


 * $\ds \map {\tilde d} {w_n, W} = \lim_{n \mathop \to \infty} \map d {w_n, W} < \epsilon$

Therefore, $\sequence {\hat w_n} \to W$ as $N \to \infty$, and $\tilde A$ is complete.