Completion Theorem (Metric Space)

Theorem
Let $(X, d)$ be a metric space.

Then there exists a completion $(\tilde X, \tilde d)$ of $(X, d)$.

Moreover, this completion is unique up to isometry.

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

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 $(X, d)$ be a metric space.

Let $\mathcal C[X]$ denote collection of all Cauchy sequences in $X$.

Define a relation $\sim$ on $\mathcal C[X]$ by


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

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

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

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

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

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

Define $\tilde d : \tilde X \to [0,\infty) \subseteq \R$ by


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

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

Proof of Lemma
Let $\left\langle x_n \right\rangle$, $\left\langle \hat x_n \right\rangle$, $\left\langle y_n \right\rangle$, $\left\langle \hat y_n \right\rangle \in \mathcal C[X]$ 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


 * $d(\hat x_n, \hat y_n) - d(x_n,y_n) \leq d(x_n,\hat x_n) + d(\hat y_n, y_n)$

and therefore,


 * $\displaystyle 0\leq \left| d(x_n,y_n) - d(\hat x_n, \hat y_n) \right| \leq d(x_n,\hat x_n) + d(\hat y_n, y_n)$

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

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

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

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

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

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

$\tilde d$ is a metric
If $\tilde d\left( [x_n], [y_n] \right) = \infty$, then $\left\langle x_n \right\rangle$ and $\left\langle y_n \right\rangle$ cannot both be Cauchy, so $\tilde d\left( [x_n], [y_n] \right) < \infty$ for $[x_n],[y_n] \in \tilde X$.

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

The closure of $[0,\infty)$ is $[0,\infty)$, so $\tilde d:\tilde X \to [0,\infty)$.

This proves Definition:positivity of $\tilde d$.

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


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

So by definition, $\left\langle x_n \right\rangle \sim \left\langle y_n \right\rangle$ and $[x_n] = [y_n]$.

This proves Definition:Definiteness.

This proves symmetry.

This proves the triangle inequality.

$\tilde X$ 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 it's isomorphic copy in $\tilde X$ when it is convenient.

For any $x, y \in X$,

So $\tilde d\big|_{X} = d$.

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 $[x_n] \in \tilde X$ and $\epsilon > 0$ be arbitrary.

If we can find $ x \in X $ such that $\tilde d( [\hat x], [x_n] ) < \epsilon$ 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 for all $m,n \geq N$, $d(x_m, x_n) < \epsilon$.

Then we have

and we are done.

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

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

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

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

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 $m,n \geq N$, we have $d(w_n, w_m) < \epsilon$.

Thus for all $n > N$,


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

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 $(X, d)$.

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

The sets $\phi_1 ( X )$ and $\phi_2 ( X ) $ are dense in $X_1$ and $X_2$ respectively, so we 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 $\displaystyle \psi( x ) = \lim_{n \to \infty} \psi (w_n)$, 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 $\psi$ is easy: for $ y \in X_2 $, let $\left\langle w_n' \right\rangle$ be a Cauchy sequence in $\phi_2 (X)$.

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

Then $\displaystyle \lim_{n \to \infty} \psi( z_n ) = y$ as required.

Injectivity of $\psi$ holds because $X_1$ and $X_2$ are Hausdorff.

Suppose that $\displaystyle \lim_{n \to \infty} \psi(w_n) = \lim_{n \to \infty} \psi(w_n')$, but $\displaystyle \lim_{n \to \infty} w_n = w$, $\displaystyle \lim_{n \to \infty} w_n' = w'$ with $w \neq w'$.

For $\epsilon > 0$, pick $M \in \N$ such that $\psi(w_n)$, $\psi(w_n')$ lie in the open ball $B_{\epsilon / 4}\left(\psi(w)\right)$ for all $n \geq M$. Then,


 * $\displaystyle \tilde d_1( w_n, w_n' ) = \tilde d_2 \left( \psi(w_n), \psi(w_n') \right) \leq \frac \epsilon 2$

Because $X_1$ and $X_2$ are Hausdorff, we can pick two disjoint neighbourhoods of $w$, $w'$, a contradiction.

Finally, because a metric space has a continuous metric, it follows that $\psi$ is an isometry on all of $X_1$, and we are done.