Metric space
Completion
- $(X, d)$,
- metric space
There is a metric space $(\bar{X}, \bar{d})$ and $\phi: X \rightarrow \bar{X}$ which satisfies
- (1) $(\bar{X}, \bar{d})$ is a complete
- (2) $\phi$ is metric preserving
- (3) $\phi(X)$ is dense in $\bar{X}$.
Moreover, such $((\bar{X}, \bar{d}), \phi)$ is unique.
Proof
(1)
Let $C(X)$ be a set of all cauchy squence. We define a equivalence relation $R$ on $C(X)$ as
\[(x_{n}) R (y_{n}) \Leftrightarrow \lim_{n}d(x_{n}, y_{n}) = 0 .\]$R$ is a equivalence relation. It is reflextive and symetric. Transivity is prooved by triangle inequality.
Let $\bar{X} := C(S) / R$ and
\[\bar{d}([(x_{n})], [(y_{n})]) := \lim_{n} d(x_{n}, y_{n}) .\]$\bar{d}$ is well-defined and metric in $\bar{X}$. Indeed, if \((x_{n})R(x^{\prime}_{n})\) and $(y_{n})R(y_{n}^{\prime})$, $\lim_{n}d(x_{n}, x_{n}^{\prime} = 0$ and $\lim_{n}d(y_{n}, y_{n}^{\prime}) = 0$. Hence
\[d(x_{n}, y_{n}) \le d(x_{n}, x_{n}^{\prime}) + d(x_{n}^{\prime}, y_{n}^{\prime}) + d(y_{n}^{\prime}, y_{n}) .\]By taking limit of both sides, we obtain
\[\lim_{n} d(x_{n}, y_{n}) \le \lim_{n} d(x_{n}^{\prime}, y_{n}^{\prime}) .\]Similary, we obtain
\[\lim_{n} d(x_{n}^{\prime}, y_{n}^{\prime}) \le \lim_{n} d(x_{n}, y_{n}) .\]By definition of $\bar{d}$, identity of indiscernibles holds. Symmetry, and triangle inequality can be derived from properties of $d$. Thus, $(\bar{X}, \bar{d})$ is a metric space.
Let $x \in X$ and $(x_{n})$ where $x_{n} := x$ for all $n \in \mathbb{N}$. $(x_{n})$ is a Cauchy sequence so we define
\[\phi(x) := [(x_{n})] .\]$\phi$ is injective. If $\phi(x) = \phi(y)$, $\lim_{n}d(x, y) = 0$. Hence $x = $.
$\phi$ is a metric preservering. For $x, y \in X$,
\[\bar{d}(\phi(x), \phi(y)) = \bar{d}([(x_{n})], [(y_{n})]) = d((x, y) .\](3) $X$ is dense in $\bar{X}$. Let $[(x_{n})] \in \bar{X}$. The sequence \((\phi(x_{n}))_{n}\) converges to $[(x_{n})]$.
\[\begin{eqnarray} \bar{d}([(x_{i})], \phi(x_{n})) & = & \lim_{i} d(x_{i}, x_{n}) \nonumber \end{eqnarray}\]$(x_{n})$ is a Cauchy sequence in $X$ so
\[\lim_{n} \bar{d}([(x_{i})], \phi(x_{n})) = 0 .\]We will show that it is complete. Let \(([(x_{n}^{i})_{n}])_{i}\) be a Cauchy sequence. $\phi(X)$ is dense in $X$. For all $i \in \mathbb{N}$, there exists $x_{i} \in X$
\[\bar{d}([(x_{n}^{i})_{n}], \phi(x_{i})) = \frac{ 1 }{ i } .\]$(x_{i})_{i}$ is a Cauchy sequence in $X$. Indeed,
\[\begin{eqnarray} \bar{d}(\phi(x_{i}), \phi(x_{j})) \le \bar{d}(\phi(x_{i}), [(x_{k}^{i})_{k}]) + \bar{d}([(x_{k}^{i})_{k}], [(x_{k}^{j})_{k}]) + \bar{d}([(x_{k}^{j})_{k}], \phi(x_{j})) \nonumber . \end{eqnarray}\]For any $\epsilon > 0$, there exists $n_{0} \in \mathbb{N}$ such that $i > n_{0}, j > n_{0}$,
\[d(x_{i}, (x_{j}) = \bar{d}(\phi(x_{i}), \phi(x_{j})) < \epsilon .\]Thus, $[(x_{i})] \in \bar{X}$.
\[\begin{eqnarray} \bar{d}([(x_{n}^{i})_{n}], [(x_{i})_{i}]) \le \bar{d}([(x_{n}^{i})_{n}], \phi(x_{i})) + \bar{d}([(x_{n}^{i})_{n}], \phi(x_{i})) \end{eqnarray}\]RHS converges to 0 as $i \rightarrow \infty$. Therefore, $(\bar{X}, \bar{d})$ is complete.
(Uniquness)