Isometric embedding of metric space

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

Theorem 9 If $langle X, ho angle$ 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 ${ ho(x_n, y_n)}_{n geq 1}$ converges.

   Proof: Because $langle X, ho angle$ 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 $ ho(x_n, y_n) in mathbb{R}$ and $mathbb{R}$ is complete, to prove ${ ho(x_n, y_n)}_{n geq 1}$ 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 $m, n > N$,

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

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

   $$ ho(x_n, y_n) leq ho(x_n, y_m) + ho(y_m, y_n) $$

   and

   $$ ho(x_n, y_m) leq ho(x_n, y_n) + ho(y_n, y_m). $$

   Hence,

   $$ ho(x_n, y_n) - ho(x_n, y_m) leq ho(y_m, y_n) $$

   and

   $$ ho(x_n, y_m) - ho(x_n, y_n) leq ho(y_m, y_n), $$

   which is actually

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

   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

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

   Then

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

   Therefore, ${ ho(x_n, y_n)}_{n geq 1}$ 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 $ ho^*({x_n}_{n geq 1}, {y_n}_{n geq 1}) = lim_{n ightarrow infty} ho(x_n, y_n)$

   Proof: Because ${ ho(x_n, y_n)}_{n geq 1}$ is convergent, the above definition of $ ho^*({x_n}_{n geq 1}, {y_n}_{n geq 1})$ is meaningful. It can be verified that $ ho^*$ satisfies the conditions of positiveness, symmetry and triangle inequality, which are derived from those of $ ho$. Then we only need to find two different Cauchy sequences, the distance between which is zero, so that $ ho^*$ 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 $ ho^*({x_n}_{n geq 1}, {y_n}_{n geq 1}) = lim_{n ightarrow infty} ho(x_n, y_n) = 0$. Therefore, $ ho^*$ 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 $ ho^* = 0$ and $X$ is isometrically embedded in $X^*$.

   Proof: According to Exercise 3 in Section 1, by letting $R := { ho^*({x_n}_{n geq 1}, {y_n}_{n geq 1}) = 0}$ 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 $ ho^*(x, y) = ho^*(x_0, y_0) = ho^*(mathcal{X}, mathcal{Y}) geq 0$. If $ ho^*(mathcal{X}, mathcal{Y}) = 0$, $ ho^*(x_0, y_0) = 0$ and for any $y$ in $mathcal{Y}$, $ ho^*(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, $ ho^*(mathcal{X}, mathcal{Y}) = 0$ implies $mathcal{X} = mathcal{Y}$. On the other hand, when $mathcal{X} = mathcal{Y}$, $ ho^*(mathcal{X}, mathcal{Y}) = ho^*(x_0, x_0) = 0$. So $ ho^*$ has the property of positive definitiveness.

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

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

   Next, let $F: X ightarrow X^*$ 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 $X$,

   $$ ho^*(F(x), F(y)) = ho^*({x, x, cdots }, {y, y, cdots}) = lim_{n ightarrow infty} ho(x, y) = ho(x, y). $$

   Meanwhile, if $F(x) = F(y)$, we have $ ho^*(F(x), F(y)) = ho(x, y) = ho(x, x) = 0$. Because $ ho$ 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 $ ho^*(F(x), F(y)) = ho(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 $ ho^*({x_n}_{n geq 1}, {x_m, x_m, cdots}) = lim_{n ightarrow 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 $langle X^*, ho^* angle$ 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 $ ho(x_{n_k}, x_{n_{k+1}}) < 2^{-k}$. This subsequence can be rewritten as ${ ilde{x}_k}_{k geq 1}$ with the condition $ ho( ilde{x}_k, ilde{x}_{k+1}) < 2^{-k}$. Then we select an arbitrary Cauchy sequence of such sequences as ${S_m}_{m geq 1}$ with $S_m = { ilde{x}_{k,m}}_{k geq 1}$ satisfying for any $varepsilon > 0$, there exists $N$ in $mathbb{N}$ such that when $m, n > N$, $ ho^*(S_m, S_n) = lim_{k ightarrow infty} ho( ilde{x}_{k,m}, ilde{x}_{k,n}) < varepsilon$. This suggests that there exists $K$ in $mathbb{N}$ such that when $k > K$, $ ho( ilde{x}_{k,k}, ilde{x}_{k,n}) < varepsilon$. This Cauchy sequence of Cauchy sequences can be illustrated as a 2-dimensional matrix with infinite length as below,

   $$ egin{pmatrix} ilde{x}_{1,1} & ilde{x}_{1,2} & ilde{x}_{1,3} & cdots \ ilde{x}_{2,1} & ilde{x}_{2,2} & ilde{x}_{2,3} & cdots \ ilde{x}_{3,1} & ilde{x}_{3,2} & ilde{x}_{3,3} & cdots \ vdots & vdots & vdots & vdots end{pmatrix}, $$

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

   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 $x = lim_{n ightarrow infty} x_n$, 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 $H circ G = { m id}_{overline{X}}$ and $G circ H = { m id}_{X^*}$. 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 $X$, $Gvert_X = F$ is isometric. Furthermore, for any $x_1$ and $x_2$ in $overline{X}$, we should prove $ ho(x_1, x_2) = ho^*({a_n}_{n geq 1}, {b_n}_{n geq 1})$ where $a_n ightarrow x_1$ and $b_n ightarrow x_2$, which is quite obvious: $ ho^*({a_n}_{n geq 1}, {b_n}_{n geq 1}) = lim_{n ightarrow infty} ho(a_n, b_n) = ho(x_1, x_2)$. Hence, $G$ is an isometry between $overline{X}$ and $X^*$.