Completion Theorem (Metric Space)

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Theorem

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


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

Moreover, this completion is unique up to isometry.


That is, if $\left({\hat A, \hat d}\right)$ is another completion of $\left({A, d}\right)$, then there is a bijection $\tau: \tilde A \leftrightarrow \hat A$ such that:

$(1): \quad \tau$ restricts to the identity on $x$:
$\forall x \in A : \tau \left({x}\right) = x$
$(2): \quad \tau$ preserves metrics:
$\forall x_1, x_2 \in A : \hat d \left({\tau \left({x_1}\right), \tau \left({x_2}\right)}\right) = \tilde d \left({x_1, x_2}\right)$


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

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

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

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

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


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 $\tilde A$.

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

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


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

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


Lemma 1: $\tilde d$ is Well-Defined

$\tilde d$ is well-defined on $\tilde A$.


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

Therefore we must show that:

$\tilde d$ is a metric on $\tilde A$
There exists an everywhere dense inclusion $\left({A, d}\right) \to \left({\tilde A, \tilde d}\right)$ preserving $d$.

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


Lemma 2: $\tilde d$ is a Metric

$\tilde d$ is a metric on $\tilde A$.


Lemma 3: $\tilde A$ is a Completion of $A$

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


Lemma 4: Uniqueness of $\tilde A$

$\tilde M = \left({\tilde A, \tilde d}\right)$ is unique up to isometry.


$\blacksquare$


Axiom of Countable Choice

This theorem depends on the Axiom of Countable Choice, by way of Completion Theorem (Metric Space)/Lemma 3.

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.


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