Completion Theorem (Metric Space)

Theorem
Let $\left({X, d}\right)$ be a metric space.

Then there exists a completion $\left({\tilde X, \tilde d}\right)$ of $\left({X, d}\right)$.

Moreover, this completion is unique up to isometry.

That is, if $\left({\hat X, \hat d}\right)$ is another completion of $\left({X, d}\right)$, then there is a bijection $\tau: \tilde X \leftrightarrow \hat X$ such that:
 * $(1): \quad \tau$ restricts to the identity on $x$:
 * $\forall x \in X : \tau \left({x}\right) = x$
 * $(2): \quad \tau$ preserves metrics:
 * $\forall x_1, x_2 \in X : \hat d \left({\tau \left({x_1}\right), \tau \left({x_2}\right)}\right) = \tilde d \left({x_1, x_2}\right)$

Proof
We construct the completion of a metric space as equivalence classes of the set of Cauchy sequences in the space under a suitable equivalence relation.

Let $\left({X, d}\right)$ be a metric space.

Let $\mathcal C \left[{X}\right]$ denote the set of all Cauchy sequences in $X$.

Define a relation $\sim$ on $\mathcal C \left[{X}\right]$ by:


 * $\displaystyle \left\langle{x_n}\right\rangle \sim \left\langle{y_n}\right\rangle \iff \lim_{n \mathop \to \infty} d \left({x_n, y_n}\right) = 0$

By Equivalence Relation on Cauchy Sequences, $\sim$ is an equivalence relation on $\mathcal C \left[{X}\right]$.

Denote the equivalence class of $\left\langle{x_n}\right\rangle \in \mathcal C \left[{X}\right]$ by $\left[{x_n}\right]$.

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

By Relation Partitions Set iff Equivalence this is a partition of $\mathcal C \left[{X}\right]$.

That is, each $\left\langle{x_n}\right\rangle \in \mathcal C \left[{X}\right]$ lies in one and only one equivalence class under $\sim$.

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


 * $\displaystyle \tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) = \lim_{n \mathop \to \infty} d \left({x_n, y_n}\right)$

Lemma
$\tilde d$ is well-defined on $\tilde X$.

Proof of Lemma
Let $\left\langle{x_n}\right\rangle$, $\left\langle{\hat x_n}\right\rangle$, $\left\langle{y_n}\right\rangle$, $\left\langle{\hat y_n}\right\rangle \in \mathcal C \left[{X}\right]$ be such that:


 * $\left\langle{x_n}\right\rangle \sim \left\langle{\hat x_n}\right\rangle$


 * $\left\langle{y_n}\right\rangle \sim \left\langle{\hat y_n}\right\rangle$

We have:

By an identical argument, we can also show that:


 * $d \left({\hat x_n, \hat y_n}\right) - d \left({x_n, y_n}\right) \le d \left({x_n, \hat x_n}\right) + d \left({\hat y_n, y_n}\right)$

and therefore:


 * $\displaystyle 0 \le \left\vert{d \left({x_n, y_n}\right) - d \left({\hat x_n, \hat y_n}\right)}\right\vert \le d \left({x_n, \hat x_n}\right) + d \left({\hat y_n, y_n}\right)$

Passing to the limit $n \to \infty$ and using the Combination Theorem for Sequences we have shown that:

$\displaystyle \lim_{n \mathop \to \infty} d \left({x_n, y_n}\right) = \lim_{n \mathop \to \infty} d \left({\hat x_n, \hat y_n}\right)$

But this precisely means that:
 * $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) = \tilde d \left({\left[{\hat x_n}\right], \left[{\hat y_n}\right]}\right)$

We claim that $\left({\tilde X, \tilde d}\right)$ is a completion of $\left({X, d}\right)$.

Therefore we must show that:
 * $\tilde d$ is a metric on $\tilde X$
 * There exists an everywhere dense inclusion $\left({X, d}\right) \to \left({\tilde X, \tilde d}\right)$ preserving $d$.

In addition the theorem claims that $\left({\tilde X, \tilde d}\right)$ is unique up to isometry.

$\tilde d$ is a metric
To prove $\tilde d$ is a metric, we verify that it satisfies the axioms M1,M2,M3 and M4.

Let $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) = \infty$.

Then $\left\langle{x_n}\right\rangle$ and $\left\langle y_n \right\rangle$ cannot both be Cauchy.

So $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) < \infty$ for $\left[{x_n}\right], \left[{y_n}\right] \in \tilde X$.

By the definition of $\tilde d$, for any $\left[{x_n}\right], \left[{y_n}\right] \in \tilde X$, $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right)$ must be a limit point of $R_{\ge 0}$.

The closure of $\R_{\ge 0}$ is $\R_{\ge 0}$, so $\tilde d: \tilde X \times \tilde X \to \R_{\ge 0}$.

This proves that $\tilde d$ satisfies M4.

Now suppose that $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) = 0$, which means that:


 * $\displaystyle \lim_{n \mathop \to \infty} d\left({x_n, y_n}\right) = 0$

So by definition:
 * $\left\langle{x_n}\right\rangle \sim \left\langle{y_n}\right\rangle$

and:
 * $\left[{x_n}\right] = \left[{y_n}\right]$

As $d$ is a metric, we also find immediately:
 * $\tilde d \left({\left[{x_n}\right], \left[{x_n}\right]}\right) = 0$

This proves that $\tilde d$ satisfies M1.

Furthermore, we have the following:

Hence, $\tilde d$ satisfies M3.

Lastly, we have:

showing $\tilde d$ also satisfies M2 and thus is a metric.

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

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

We first show that $\phi$ is an injection of $X$ into $\tilde X$.

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

For any $x, y \in X$,

So $\tilde d \big\vert_X = d$.

Next we show that $X$ is dense in $\tilde X$.

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

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

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

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 $X$ is dense in $\tilde X$.

Finally we must show that $\left({\tilde X, \tilde d}\right)$ is complete.

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

Let $\left\langle{\hat w_n}\right\rangle$ be a Cauchy sequence in $\phi \left({X}\right)$, 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 $X$.

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

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

Let $\epsilon > 0$ be arbitrary.

Since $\left\langle{w_1, w_2, w_3, \ldots}\right\rangle$ is Cauchy in $X$, 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 $\tilde X$ is complete.

Uniqueness of $\tilde X$
Suppose that $\left({\tilde{X_1}, \tilde{d_1}, \phi_1}\right)$, $\left({\tilde{X_2}, \tilde{d_2}, \phi_2}\right)$ are two completions of $\left({X, d}\right)$.

Then $\psi = \phi_1^{-1} \circ \phi_2$ gives an isometry from $\phi_1 \left({X}\right)$ to $\phi_2 \left({X}\right)$.

The sets $\phi_1 \left({X}\right)$ and $\phi_2 \left({X}\right)$ are dense in $X_1$ and $X_2$ respectively.

Thus we can extend $\psi$ continuously to a map $\psi: X_1 \to X_2$.

That is, for $x \in X_1$, we can find a Cauchy sequence $\left\langle{w_n}\right\rangle$ in $X_1$ with limit $x$.

Then we define:
 * $\displaystyle \psi \left({x}\right) := \lim_{n \mathop \to \infty} \psi \left({w_n}\right)$

which converges as $X_2$ is complete.

By Metric Space is Hausdorff, $X_1$ and $X_2$ are Hausdorff.

Therefore, by Convergent Sequence in Hausdorff Space has Unique Limit, $\psi$ is well defined.

Surjectivity of $\psi$:

For $y \in \tilde {X_2}$, let $\left\langle{w_n'}\right\rangle$ be a Cauchy sequence in $\phi_2 \left({X}\right)$ with limit $y$ in $\tilde{X_2}$.

Let $z_n$ be the preimage of the $w_n'$ under $\psi$.

Then, as $X_2$ is Hausdorff:
 * $\displaystyle \lim_{n \mathop \to \infty} \psi \left({z_n}\right) = y$

as required.

injectivity of $\psi$ holds because $X_1$ is Hausdorff, as follows:

Suppose that:
 * $\displaystyle \lim_{n \mathop \to \infty} \psi \left({w_n}\right) = \lim_{n \mathop \to \infty} \psi \left({w_n'}\right)$
 * $\displaystyle \lim_{n \mathop \to \infty} w_n = w$

and:
 * $\displaystyle \lim_{n \mathop \to \infty} w_n' = w'$.

For $\epsilon > 0$, pick $M \in \N$ such that $\psi \left({w_n}\right)$, $\psi \left({w_n'}\right)$ lie in the open $\epsilon$-ball $B_{\epsilon / 2} \left({\psi \left({w}\right)}\right)$ for all $n \ge M$.

Then we have:


 * $\tilde {d_1} \left({w_n, w_n'}\right) = \tilde {d_2} \left({\psi \left({w_n}\right), \psi \left({w_n'}\right)}\right) \le \epsilon$

As $X_1$ is Hausdorff, we conclude $w = w'$, and the task is complete.

Finally, from Metric is Continuous, it follows that $\psi$ is an isometry on all of $X_1$, and the proof is complete.