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 \(X^*\).

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