Completion Theorem (Metric Spaces)
Contents |
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
- $\tau$ restricts to the identity on $x$:
- $\forall x \in X : \tau \left({x}\right) = x$
- $\tau$ 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 collection of all Cauchy sequences in $X$.
Define a relation $\sim$ on $\mathcal C \left[{X}\right]$ by:
- $\displaystyle \left\langle x_n \right\rangle \sim \left\langle y_n \right\rangle \iff \lim_{n \to \infty} d \left({x_n, y_n}\right) = 0$
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 a 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 $\tilde d : \tilde X \to [0,\infty) \subseteq \R$ by:
- $\displaystyle \tilde d \left({ \left[{x_n}\right], \left[{y_n}\right]}\right) = \lim_{n \to \infty} d \left({x_n, y_n}\right)$
Lemma
$\tilde d$ is well-defined on $\tilde X$.
Proof of Lemma
Let $\left\langle x_n \right\rangle$, $\left\langle \hat x_n \right\rangle$, $\left\langle y_n \right\rangle$, $\left\langle \hat y_n \right\rangle \in \mathcal C[X]$ 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
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle d(x_n,y_n) - d(\hat x_n, \hat y_n)\) | \(\leq\) | \(\displaystyle d(x_n,\hat x_n) + d(\hat x_n,y_n) - d(\hat x_n, \hat y_n)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | by the triangle inequality | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\leq\) | \(\displaystyle d(x_n,\hat x_n) + d(\hat x_n, \hat y_n) + d(\hat y_n, y_n) - d(\hat x_n, \hat y_n)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | by the triangle inequality | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle d(x_n,\hat x_n) + d(\hat y_n, y_n)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
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 $\tilde d \left({ [x_n], [y_n] }\right) = \tilde d \left({ [\hat x_n], [\hat y_n] }\right)$.
$\Box$
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 an everywhere 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
To prove $\tilde d$ is a metric, we verify that it satisfies the axioms M1',M2,M3 and M4'.
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 $\tilde d:\tilde X\times\tilde X \to [0,\infty)$.
This proves that $\tilde d$ satisfies M4'.
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 $[x_n] = [y_n]$.
As $d$ is a metric, we also find immediately $\tilde d\left( [x_n], [x_n] \right)=0$.
This proves that $\tilde d$ satisfies M1'.
Furthermore, we have the following:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \tilde d\left( [x_n], [y_n] \right)\) | \(=\) | \(\displaystyle \lim_{n \to \infty} d\left( x_n, y_n \right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{n \to \infty} d\left( y_n, x_n \right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | because $d$ is a metric | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \tilde d\left( [y_n], [x_n] \right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
Hence, $\tilde d$ satisfies M3.
Lastly, we have:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \tilde d \left( [x_n], [z_n] \right)\) | \(=\) | \(\displaystyle \lim_{n \to \infty} d\left( x_n, z_n \right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\leq\) | \(\displaystyle \lim_{n \to \infty} \left\{ d\left( x_n, y_n \right) + d\left( y_n, z_n \right) \right\}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | because $d$ is a metric, and using elementary properties of limits (Reference?) | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{n \to \infty} d\left( x_n, y_n \right) + \lim_{n \to \infty} d\left( y_n, z_n \right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | sum rule for limits | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \tilde d \left( [x_n], [y_n] \right) + \tilde d \left( [y_n], [z_n] \right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
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 $\tilde X$.
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \phi(x) = \phi(y)\) | \(\implies\) | \(\displaystyle \left[ \hat x \right] = \left[ \hat y \right]\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle \lim_{n \to \infty} d(x, y) = 0\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle d(x, y) = 0\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle x = y\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
Henceforth we identify $X$ with its isomorphic copy in $\tilde X$ when it is convenient.
For any $x, y \in X$,
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \tilde d( [\hat x], [\hat y])\) | \(=\) | \(\displaystyle \lim_{n \to \infty} d(x, y)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle d(x,y)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
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:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \tilde d ( [\hat x_N], [ x_n ] )\) | \(=\) | \(\displaystyle \lim_{n \to \infty} d( x_N, x_n )\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(<\) | \(\displaystyle \epsilon\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
and therefore $X$ is dense in $\tilde X$.
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 \tilde{X_2} $, let $\left\langle w_n' \right\rangle$ be a Cauchy sequence in $\phi_2 (X)$ 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, $\displaystyle \lim_{n \to \infty} \psi( z_n ) = y$ as required.
Injectivity of $\psi$ holds because $X_1$ is Hausdorff.
Suppose that $\displaystyle \lim_{n \to \infty} \psi(w_n) = \lim_{n \to \infty} \psi(w_n')$, $\displaystyle \lim_{n \to \infty} w_n = w$, and $\displaystyle \lim_{n \to \infty} w_n' = w'$.
For $\epsilon > 0$, pick $M \in \N$ such that $\psi(w_n)$, $\psi(w_n')$ lie in the open ball $B_{\epsilon / 2}\left(\psi(w)\right)$ for all $n \geq M$. Then we have:
- $\displaystyle \tilde{d_1}( w_n, w_n' ) = \tilde{d_2} \left( \psi(w_n), \psi(w_n') \right) \leq \epsilon$
As $X_1$ is Hausdorff, we conclude $w=w'$, and we are done.
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.
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Axiom of Countable Choice
This theorem depends on the Axiom of Countable Choice, by way of Completeness Criterion (Metric Spaces).
Although not as strong as the Axiom of Choice, the Axiom of Countable Choice is similarly independent of the Zermelo-Fraenkel axioms.
As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.
In particular: Not many, but they need to be reviewed..
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Sources
- Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 5$: Complete Metric Spaces
