Isometric embedding of metric space

This post summarizes the proof for Theorem 7.9 in Royden's "Real Analysis".

Theorem 9 If $\langle X, \rho \rangle$ is an incomplete metric space, it is possible to find a complete metric space $X^*$ in which $X$ is isometrically embedded as a dense subset. If $X$ is contained in an arbitrary complete space $Y$, then $X^*$ is isometric with the closure of $X$ in $Y$.

Analysis: The proof about the completion of $X$ is not straightforward because this problem is given in an abstract way that there is nothing to manipulate or we cannot construct a specific example as the completion of $X$ for facilitating our understanding. Hence a new structure must be introduced to fulfill this purpose, which is the set of equivalence classes $X^*$ derived from the collection of all Cauchy sequences in $X$. Then it is to be proved that $X$ is identified with $F(X)$ in $X^*$, where $F$ is an isometry from $X$ to $F(X)$ and $F(X)$ is dense in $X^*$.

The proof of this theorem is divided into 5 steps as suggested by Exercise 17.

1. If $\{x_n\}_{n \geq 1}$ and $\{y_n\}_{n \geq 1}$ are Cauchy sequences from a metric space $X$, then $\{\rho(x_n, y_n)\}_{n \geq 1}$ converges.

   Proof: Because $\langle X, \rho \rangle$ is an incomplete space, the two sequences $\{x_n\}_{n \geq 1}$ and $\{y_n\}_{n \geq 1}$ do not necessarily have their limits within $X$. However, this proposition indicates that the inter-distance between $x_n$ and $y_n$ does converge when $n$ approaches to $\infty$.

   Because $\rho(x_n, y_n) \in \mathbb{R}$ and $\mathbb{R}$ is complete, to prove $\{\rho(x_n, y_n)\}_{n \geq 1}$ converges, we need to show that it is a Cauchy sequence.

   From the given condition, for any $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that when $m_1, m_2 > N$, $\abs{x_{m_1} - x_{m_2}} < \frac{\varepsilon}{2}$ and when $n_1, n_2 > N$, $\abs{y_{n_1} - y_{n_2}} < \frac{\varepsilon}{2}$.

   For any $m, n > N$,

   $$ \abs{\rho(x_n, y_n) - \rho(x_m ,y_m)} = \abs{\rho(x_n, y_n) - \rho(x_n, y_m) + \rho(x_n, y_m) - \rho(x_m, y_m)}. $$

   Due to the triangle inequality satisfied by the metric $\rho$, we have

   $$ \rho(x_n, y_n) \leq \rho(x_n, y_m) + \rho(y_m, y_n) $$

   and

   $$ \rho(x_n, y_m) \leq \rho(x_n, y_n) + \rho(y_n, y_m). $$

   Hence,

   $$ \rho(x_n, y_n) - \rho(x_n, y_m) \leq \rho(y_m, y_n) $$

   and

   $$ \rho(x_n, y_m) - \rho(x_n, y_n) \leq \rho(y_m, y_n), $$

   which is actually

   $$ \abs{\rho(x_n, y_n) - \rho(x_n, y_m)} \leq \rho(y_m, y_n). $$

   This is just a variation of the triangle inequality for a metric which says that the difference between the lengths of two edges of a triangle is smaller than or equal to the length of the third edge. Similarly, we have

   $$ \abs{\rho(x_n, y_m) - \rho(x_m, y_m)} \leq \rho(x_m, x_n). $$

   Then

   $$ \begin{aligned} \abs{\rho(x_n, y_n) - \rho(x_m, y_m)} &= \abs{\rho(x_n, y_n) - \rho(x_n, y_m) + \rho(x_n, y_m) - \rho(x_m, y_m)} \\ & \leq \abs{\rho(x_n, y_n) - \rho(x_n, y_m)} + \abs{\rho(x_n, y_m) - \rho(x_m, y_m)} \\ & \leq \rho(y_m, y_n) + \rho(x_m, x_n) \\ & < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\ & = \varepsilon. \end{aligned} $$

   Therefore, $\{\rho(x_n, y_n)\}_{n \geq 1}$ is a Cauchy sequence in $\mathbb{R}$ and converges to some $a$ in $\mathbb{R}$.

2. The set of all Cauchy sequences from a metric space $X$ becomes a pseudometric space if $\rho^*(\{x_n\}_{n \geq 1}, \{y_n\}_{n \geq 1}) = \lim_{n \rightarrow \infty} \rho(x_n, y_n)$

   Proof: Because $\{\rho(x_n, y_n)\}_{n \geq 1}$ is convergent, the above definition of $\rho^*(\{x_n\}_{n \geq 1}, \{y_n\}_{n \geq 1})$ is meaningful. It can be verified that $\rho^*$ satisfies the conditions of positiveness, symmetry and triangle inequality, which are derived from those of $\rho$. Then we only need to find two different Cauchy sequences, the distance between which is zero, so that $\rho^*$ is a pseudometric.

   Let $\{z_k\}_{k \geq 1}$ be a Cauchy sequence. Let $x_n = z_{2n-1}$ and $y_n = z_{2n}$. Then $\{x_n\}_{n \geq 1}$ and $\{y_n\}_{n \geq 1}$ are two different Cauchy sequences with $\rho^*(\{x_n\}_{n \geq 1}, \{y_n\}_{n \geq 1}) = \lim_{n \rightarrow \infty} \rho(x_n, y_n) = 0$. Therefore, $\rho^*$ defined on the collection of all Cauchy sequences in $X$ is a pseudometric.

3. This pseudometric space becomes a metric space $X^*$ when we identify elements for which $\rho^* = 0$ and $X$ is isometrically embedded in $X^*$.

   Proof: According to Exercise 3 in Section 1, by letting $R := \{\rho^*(\{x_n\}_{n \geq 1}, \{y_n\}_{n \geq 1}) = 0\}$ be the equivalence condition on the set of all Cauchy sequences in $X$, the obtained collection of equivalence classes $X^*$ is a metric space. This can be verified as below.

   Let $\mathcal{X}$ and $\mathcal{Y}$ be two different equivalence classes in $X^*$. Let $x_0$ be the representative element of $\mathcal{X}$ and $y_0$ be that of $\mathcal{Y}$. Then for any $x$ in $\mathcal{X}$ and any $y$ in $\mathcal{Y}$, we have $\rho^*(x, y) = \rho^*(x_0, y_0) = \rho^*(\mathcal{X}, \mathcal{Y}) \geq 0$. If $\rho^*(\mathcal{X}, \mathcal{Y}) = 0$, $\rho^*(x_0, y_0) = 0$ and for any $y$ in $\mathcal{Y}$, $\rho^*(x_0, y) = 0$. Because of the equivalence relation $R$, $y$ belongs to $\mathcal{X}$. Similarly, for any $x$ in $\mathcal{X}$, $x$ belongs to $\mathcal{Y}$. Therefore, $\rho^*(\mathcal{X}, \mathcal{Y}) = 0$ implies $\mathcal{X} = \mathcal{Y}$. On the other hand, when $\mathcal{X} = \mathcal{Y}$, $\rho^*(\mathcal{X}, \mathcal{Y}) = \rho^*(x_0, x_0) = 0$. So $\rho^*$ has the property of positive definitiveness.

   The commutativity of $\rho^*$ is obvious, which is derived from that of $\rho$.

   Finally, for $\mathcal{X}$, $\mathcal{Y}$, $\mathcal{Z}$ in $X^*$ with their respective representative elements $x_0$, $y_0$ and $z_0$, $\rho(\mathcal{X}, \mathcal{Y}) = \rho(x_0, y_0)$, $\rho(\mathcal{X}, \mathcal{Z}) = \rho(x_0, z_0)$ and $\rho(\mathcal{Z}, \mathcal{Y}) = \rho(z_0, y_0)$. Because $\rho(x_0, y_0) \leq \rho(x_0, z_0) + \rho(z_0, y_0)$, we have the triangle inequality for $\rho^*$. Therefore, $X^*$ with $\rho^*$ is a metric space.

   Next, let $F: X \rightarrow X^*$ which associates each $x$ in $X$ with the equivalence class in $X^*$ that contains the Cauchy sequence $\{x, x, \cdots\}$. It is easy to see that for any $x$, $y$ in $X$,

   $$ \rho^*(F(x), F(y)) = \rho^*(\{x, x, \cdots \}, \{y, y, \cdots\}) = \lim_{n \rightarrow \infty} \rho(x, y) = \rho(x, y). $$

   Meanwhile, if $F(x) = F(y)$, we have $\rho^*(F(x), F(y)) = \rho(x, y) =\rho(x, x) = 0$. Because $\rho$ is a standard metric, $x = y$. Hence, $F$ is injective. For any open ball $B(\{x, x, \cdots\}, \varepsilon)$ with a radius $\varepsilon$ in $F(X)$, its inverse image under $F^{-1}$ is $B(x, \varepsilon)$ in $X$, which is open in $X$. Then $F$ is a continuous map. Similarly, $F^{-1}$ is also continuous. Therefore, $F$ is a homeomorphism between $X$ and $F(X)$. Moreover, because $\rho^*(F(x), F(y)) = \rho(x, y)$, $F$ is an isometry.

   Then, we will prove $X$ is isometrically embedded in $X^*$ as a dense subset.

   Let $B(\{x_n\}_{n \geq 1}, \varepsilon)$ be an open ball in $X^*$, which is centered at an arbitrary element $\{x_n\}_{n \geq 1}$ in $X^*$. Because $\{x_n\}_{n \geq 1}$ is a Cauchy sequence, there exists $N$ in $\mathbb{N}$ such that when $m, n > N$, $\abs{x_m - x_n} < \varepsilon$. Then $\rho^*(\{x_n\}_{n \geq 1}, \{x_m, x_m, \cdots\}) = \lim_{n \rightarrow \infty} \abs{x_n - x_m} < \varepsilon$. Hence $\{x_m, x_m, \cdots\}$ belongs to $B(\{x_n\}_{n \geq 1}, \varepsilon)$ and $F(X)$ is dense in $X^*$. Because $F$ is an isometry from $X$ to $F(X)$, $X$ can be identified with $F(X)$. Therefore, $X$ is isometrically embedded in $X^*$.

4. The metric space $\langle X^*, \rho^* \rangle$ is complete. (N.B. What is convergent here is a sequence of Cauchy sequences.)

   Proof: Let $\{x_n\}_{n \geq 1}$ be any Cauchy sequence in $X$. We can extract a subsequence from it as $\{x_{n_k}\}_{k \geq 1}$ such that $\rho(x_{n_k}, x_{n_{k+1}}) < 2^{-k}$. This subsequence can be rewritten as $\{\tilde{x}_k\}_{k \geq 1}$ with the condition $\rho(\tilde{x}_k, \tilde{x}_{k+1}) < 2^{-k}$. Then we select an arbitrary Cauchy sequence of such sequences as $\{S_m\}_{m \geq 1}$ with $S_m = \{\tilde{x}_{k,m}\}_{k \geq 1}$ satisfying for any $\varepsilon > 0$, there exists $N$ in $\mathbb{N}$ such that when $m, n > N$, $\rho^*(S_m, S_n) = \lim_{k \rightarrow \infty} \rho(\tilde{x}_{k,m}, \tilde{x}_{k,n}) < \varepsilon$. This suggests that there exists $K$ in $\mathbb{N}$ such that when $k > K$, $\rho(\tilde{x}_{k,k}, \tilde{x}_{k,n}) < \varepsilon$. This Cauchy sequence of Cauchy sequences can be illustrated as a 2-dimensional matrix with infinite length as below,

   $$ \begin{pmatrix} \tilde{x}_{1,1} & \tilde{x}_{1,2} & \tilde{x}_{1,3} & \cdots \\ \tilde{x}_{2,1} & \tilde{x}_{2,2} & \tilde{x}_{2,3} & \cdots \\ \tilde{x}_{3,1} & \tilde{x}_{3,2} & \tilde{x}_{3,3} & \cdots \\ \vdots & \vdots & \vdots & \vdots \end{pmatrix}, $$

   from which we extract the diagonal elements to construct a new sequence $S^* = \{\tilde{x}_{k,k}\}_{k \geq 1}$. For any $\varepsilon > 0$, there exists $N$ in $\mathbb{N}$ such that when $m, n > N$, $\rho(\tilde{x}_{m,m}, \tilde{x}_{n,n}) < \varepsilon$. Hence $S^*$ is a Cauchy sequence so it belongs to $X^*$. The distance between $S_m$ and $S^*$ is $\rho^*(S_m, S^*) = \lim_{k \rightarrow \infty} \rho(\tilde{x}_{k,m}, \tilde{x}_{k,k})$. It is obvious that for any $\varepsilon > 0$, when $m > N$ and $k > K$, $\rho(\tilde{x}_{k,m}, \tilde{x}_{k,k}) < \varepsilon$. Therefore, $\lim_{m \rightarrow \infty} \rho^*(S_m, S^*) = 0$ and $\langle X^*, \rho^* \rangle$ is complete.

   Up to now, the first part of Theorem 9 is proved, i.e. we have found the completion of $X$ as $X^*$ in which $X$ is isometrically embedded as a dense subset.

5. The above isometry $F$ from $X$ to $F(X)$ in $X^*$ is uniformly continuous, which is because for any $x$ and $y$ in $X$ such that $\rho(x, y) < \varepsilon$, $\rho^*(F(x), F(y)) = \lim_{n \rightarrow \infty} \rho(x, y) = \rho(x, y) < \varepsilon$. Then according to Proposition 11 in Section 5 of this Chapter, viewing $X$ as a subset of $Y$, $F$ is a uniformly continuous mapping from $X$ into the complete space $X^*$. Then there exists a unique continuous extension $G$ of $F$ from $X$ to $\overline{X}$ with $\overline{X}$ being the closure of $X$ with respect to the standard topology induced by the metric. Because $Y$ is also complete with respect to this topology, $\overline{X}$ is contained within $Y$. Also note that, for any $x$ in $\overline{X}$ but not in $X$, there exists a Cauchy sequence $\{x_n\}_{n \geq 1}$ in $X$ convergent to $x$. Then the value $G(x)$ only depends on $x$, i.e. $G(x) = \lim_{n \rightarrow \infty} F(x_n)$.

   Due to Proposition 10 in Section 5, when $\{x_n\}_{n \geq 1}$ is a Cauchy sequence in $X$, $\{F(x_n)\}_{n \geq 1}$ is also a Cauchy sequence because $F$ is uniformly continuous. Therefore, $G(x)$ belongs to the closure of $F(X)$ in $X^*$, i.e. $\overline{F(X)}$. Meanwhile, for any $y$ in $\overline{F(X)}$, there exists a Cauchy sequence $\{y_n\}_{n \geq 1}$ in $F(X)$ and $\{x_n\}_{n \geq 1}$ in $X$ such that $x_n = F^{-1}(y_n)$. Because $F$ is an isometry from $X$ to $F(X)$, $F^{-1}$ is an isometry from $F(X)$ to $X$ and hence $F^{-1}$ is also uniformly continuous. Therefore, $\{x_n\}_{n \geq 1}$ is a Cauchy sequence in $X$. Then, according to the definition of $G$, let $x$ in $\overline{X}$ and $x = \lim_{n \rightarrow \infty} x_n$, we have $G(x) = y$. This means that the actual range of $G$ is $\overline{F(X)} = X^*$.

   On the other hand, viewing $F(X)$ as a subset of $X^*$, $F^{-1}$ is an isometry from $F(X)$ to $X \subset \overline{X}$, which is also uniformly continuous. Then there exists a unique extension $H$ of $F^{-1}$ from $F(X)$ to $\overline{F(X)} = X^*$. So $H$ is a map from $X^*$ into $Y$. With a similar analysis as that for $G$, the actual range of $H$ is $\overline{X}$.

   Then we have $H \circ G = {\rm id}_{\overline{X}}$ and $G \circ H = {\rm id}_{X^*}$. Therefore, $G$ is the inverse of $H$ and vice versa. Because $G$ is uniformly continuous, $G$ is a homeomorphism. Then we need to prove $G$ is isometric. We already know that when $G$ constrained to $X$, $G\vert_X = F$ is isometric. Furthermore, for any $x_1$ and $x_2$ in $\overline{X}$, we should prove $\rho(x_1, x_2) = \rho^*(\{a_n\}_{n \geq 1}, \{b_n\}_{n \geq 1})$ where $a_n \rightarrow x_1$ and $b_n \rightarrow x_2$, which is quite obvious: $\rho^*(\{a_n\}_{n \geq 1}, \{b_n\}_{n \geq 1}) = \lim_{n \rightarrow \infty} \rho(a_n, b_n) = \rho(x_1, x_2)$. Hence, $G$ is an isometry between $\overline{X}$ and $X^*$.