Proof of The Metric Completion is a Complete Metric Space with a Dense Isometric Copy of the Space, and Maps Preserving Cauchy Sequences Extend to It

Each result cited below is universally quantified over the data in its own statement.

Preliminaries. (i) Conventions. The four conditions of Metric Space are referred to as conditions 1 to 4. For a real $\varepsilon>0$ write $\varepsilon/2=\varepsilon\cdot2^{-1}$ and $\varepsilon/4=(\varepsilon/2)/2$; by clause 8 of Elementary Order Arithmetic in an Ordered Field these are positive, $\varepsilon/2+\varepsilon/2=\varepsilon$ and $\varepsilon/4+\varepsilon/4=\varepsilon/2$. A sum of inequalities of which one is strict is strict by clause 3 of Elementary Order Arithmetic in an Ordered Field, and a chain mixing $\le$ and $<$ is strict by clause 2 there. Given $N_{1},N_{2}\in\mathbb{N}$, one of $N_{1}<N_{2}$, $N_{1}=N_{2}$, $N_{2}<N_{1}$ holds by clause 3 of Properties of the Order on the Natural Numbers; letting $N$ be the larger of the two, every $k\ge N$ satisfies $k\ge N_{1}$ and $k\ge N_{2}$ by clause 1 there. We write $\xi_{m}\to\xi$ for convergence in a metric space.

(ii) Real limits. A real sequence $(a_{k})_{k\in\mathbb{N}}$ converges to $L$ in the sense of Limit of a Sequence of Real Numbers if and only if it converges to $L$ in the real line $(\mathbb{R},d_{\mathbb{R}})$ of The Absolute Value Metric on the Real Line, since both conditions require $|a_{k}-L|<\varepsilon$ for all large $k$ and $d_{\mathbb{R}}(a_{k},L)=|a_{k}-L|$. By Uniqueness of Limits in a Metric Space, limits in any metric space, and hence limits of real sequences, are unique.

(iii) Constant sequences. In a metric space $(Z,\rho)$ a constant sequence with value $c$ converges to $c$, since $\rho(c,c)=0<\varepsilon$ for every $\varepsilon>0$ by condition 2.

(iv) Continuity of the distance. Let $(Z,\rho)$ be a metric space. For $p,q,p',q'\in Z$ put $c=\rho(p,p')+\rho(q,q')$. Conditions 4 and 3 give $\rho(p,q)\le\rho(p,p')+\rho(p',q')+\rho(q',q)=\rho(p',q')+c$, that is $\rho(p,q)-\rho(p',q')\le c$ by clause 3 of Elementary Arithmetic in an Ordered Field; exchanging $(p,q)$ with $(p',q')$, using condition 3 to rewrite $\rho(p',p)+\rho(q',q)$ as $c$, and applying clause 4 of Elementary Order Arithmetic in an Ordered Field gives $-c\le\rho(p,q)-\rho(p',q')$. By clause 6 of Properties of the Absolute Value in an Ordered Field,

Consequently, if $p_{k}\to p$ and $q_{k}\to q$ in $(Z,\rho)$, then the real sequence $(\rho(p_{k},q_{k}))_{k\in\mathbb{N}}$ converges to $\rho(p,q)$: given $\varepsilon>0$, choose $K_{1}$ with $\rho(p_{k},p)<\varepsilon/2$ for $k\ge K_{1}$, then $K_{2}$ with $\rho(q_{k},q)<\varepsilon/2$ for $k\ge K_{2}$, and let $K$ be the larger; for $k\ge K$, (P) gives $|\rho(p_{k},q_{k})-\rho(p,q)|<\varepsilon$.

(v) A null sequence of radii. Let $\iota$ be the canonical map of $\mathbb{R}$ and $r_{m}=\iota(m)^{-1}$ for $m\in\mathbb{N}$. By clause 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, $r_{m}$ exists and $0<r_{m}$. For every $\varepsilon>0$ there is $M\in\mathbb{N}$ with $r_{m}<\varepsilon$ for all $m\ge M$: clause 3 of The Archimedean Property of the Real Numbers gives $M$ with $\iota(M)^{-1}<\varepsilon$; for $m\ge M$ we have $\iota(M)\le\iota(m)$ (equality if $m=M$, clause 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field if $M<m$), so $r_{m}\le\iota(M)^{-1}<\varepsilon$ by Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares §reciprocal.

(vi) Elements of the completion. By The Metric Completion of a Metric Space §completion, every element of $\widehat{X}$ equals $[x]$ for some $x\in\mathcal{C}(X)$, and $\widehat{d}([x],[y])=\delta(x,y)$ for $x,y\in\mathcal{C}(X)$, where $\delta$ is the limit distance of Cauchy Sequences in a Metric Space: Constant Sequences, Convergence of Termwise Distances, the Null Relation, and Independence of Representatives §distances. By The Metric Completion of a Metric Space §canonical-map, $\kappa_{X}(a)=[\bar{a}]$, where $\bar{a}\in\mathcal{C}(X)$ by Cauchy Sequences in a Metric Space: Constant Sequences, Convergence of Termwise Distances, the Null Relation, and Independence of Representatives §constant.

(vii) Sequential continuity. Once clause 1 is proved, for a metric space $(Z,\rho)$ and a map $\varphi:\widehat{X}\to Z$ we apply Continuity Between Metric Spaces is Equivalent to Sequential Continuity §on-subset with $A=\widehat{X}$, whose restricted metric is $\widehat{d}$ itself: $\varphi$ is continuous if and only if $\varphi(\xi_{m})\to\varphi(\xi)$ in $(Z,\rho)$ whenever $\xi_{m}\to\xi$ in $(\widehat{X},\widehat{d})$.

Proof of clause 1 (Metric). By The Metric Completion of a Metric Space §completion, $\widehat{d}$ is a well-defined map $\widehat{X}\times\widehat{X}\to\mathbb{R}$. Let $\alpha,\beta,\gamma\in\widehat{X}$ and by (vi) write $\alpha=[x]$, $\beta=[y]$, $\gamma=[z]$ with $x,y,z\in\mathcal{C}(X)$. By Cauchy Sequences in a Metric Space: Constant Sequences, Convergence of Termwise Distances, the Null Relation, and Independence of Representatives §distances, $\widehat{d}(\alpha,\beta)=\delta(x,y)\ge0$, $\widehat{d}(\alpha,\beta)=\delta(y,x)=\widehat{d}(\beta,\alpha)$ and $\widehat{d}(\alpha,\gamma)=\delta(x,z)\le\delta(x,y)+\delta(y,z)=\widehat{d}(\alpha,\beta)+\widehat{d}(\beta,\gamma)$; these are conditions 1, 3 and 4. For condition 2: $\widehat{d}(\alpha,\beta)=0$ means $\delta(x,y)=0$, that is $x\approx y$, which by Cauchy Sequences in a Metric Space: Constant Sequences, Convergence of Termwise Distances, the Null Relation, and Independence of Representatives §null holds if and only if $[x]=[y]$, that is $\alpha=\beta$.

Proof of clause 2 (Isometry). Let $a,b\in X$. By (vi) and the last identity of Cauchy Sequences in a Metric Space: Constant Sequences, Convergence of Termwise Distances, the Null Relation, and Independence of Representatives §distances, $\widehat{d}(\kappa_{X}(a),\kappa_{X}(b))=\widehat{d}([\bar{a}],[\bar{b}])=\delta(\bar{a},\bar{b})=d(a,b)$. If $\kappa_{X}(a)=\kappa_{X}(b)$, then $d(a,b)=\widehat{d}(\kappa_{X}(a),\kappa_{X}(a))=0$ by condition 2 for $\widehat{d}$ (clause 1), so $a=b$ by condition 2 for $d$. Thus $\kappa_{X}$ is injective.

Proof of clause 3 (Density). Let $x\in\mathcal{C}(X)$ and $\varepsilon>0$. First choose $N$ by Cauchy Sequence in a Metric Space with $d(x_{k},x_{l})<\varepsilon/2$ for all $k,l\ge N$. Now fix $k\ge N$. By (vi), $\kappa_{X}(x_{k})=[\overline{x_{k}}]$, so $\widehat{d}(\kappa_{X}(x_{k}),[x])=\delta(\overline{x_{k}},x)=:L_{k}$, which by Cauchy Sequences in a Metric Space: Constant Sequences, Convergence of Termwise Distances, the Null Relation, and Independence of Representatives §distances is the limit of the real sequence $(d(x_{k},x_{l}))_{l\in\mathbb{N}}$. Next choose $N'$ with $|d(x_{k},x_{l})-L_{k}|<\varepsilon/2$ for all $l\ge N'$, and let $l$ be the larger of $N$ and $N'$. By clauses 2 and 3 of Properties of the Absolute Value in an Ordered Field, $L_{k}-d(x_{k},x_{l})\le|d(x_{k},x_{l})-L_{k}|<\varepsilon/2$, hence $L_{k}<d(x_{k},x_{l})+\varepsilon/2<\varepsilon$. Thus $\widehat{d}(\kappa_{X}(x_{k}),[x])<\varepsilon$ for every $k\ge N$, and $(\kappa_{X}(x_{k}))_{k\in\mathbb{N}}$ converges to $[x]$ by Convergent Sequence in a Metric Space.

For density, let $\mathcal{T}$ be the collection of open subsets of $(\widehat{X},\widehat{d})$, a topology by Metric Open Sets Form a Topology. The closure $\operatorname{cl}_{\widehat{X}}(\kappa_{X}(X))$ is by definition a subset of $\widehat{X}$. Conversely, let $\xi\in\widehat{X}$ and write $\xi=[x]$ by (vi). The sequence $(\kappa_{X}(x_{k}))_{k\in\mathbb{N}}$ has all its terms in $\kappa_{X}(X)$ and converges to $\xi$ by the first part, so $\xi\in\operatorname{cl}_{\widehat{X}}(\kappa_{X}(X))$ by Sequential Characterization of the Closure in a Metric Space. Hence $\operatorname{cl}_{\widehat{X}}(\kappa_{X}(X))=\widehat{X}$, which is density in the sense of Dense Subset of a Topological Space.

Proof of clause 4 (Completeness). Let $(\xi_{m})_{m\in\mathbb{N}}$ be a Cauchy sequence in $(\widehat{X},\widehat{d})$. For $m\in\mathbb{N}$ let $A_{m}=\{a\in X:\ \widehat{d}(\kappa_{X}(a),\xi_{m})<r_{m}\}$, with $r_{m}$ from (v). Each $A_{m}$ is nonempty: writing $\xi_{m}=[x]$ by (vi), clause 3 gives $\kappa_{X}(x_{k})\to\xi_{m}$, and since $0<r_{m}$ some $k$ satisfies $\widehat{d}(\kappa_{X}(x_{k}),\xi_{m})<r_{m}$, so $x_{k}\in A_{m}$. By the axiom of countable choice, applied to the family $(A_{m})_{m\in\mathbb{N}}$ of subsets of $X$, there is a sequence $a=(a_{m})_{m\in\mathbb{N}}$ in $X$ with $a_{m}\in A_{m}$ for every $m$.

The sequence $a$ is Cauchy in $(X,d)$. For $m,l\in\mathbb{N}$, clause 2 and conditions 3 and 4 for $\widehat{d}$ give

Given $\varepsilon>0$, first choose $M_{1}$ by (v) with $r_{m}<\varepsilon/4$ for $m\ge M_{1}$, then $M_{2}$ with $\widehat{d}(\xi_{m},\xi_{l})<\varepsilon/2$ for $m,l\ge M_{2}$, and let $M$ be the larger. For $m,l\ge M$ we get $d(a_{m},a_{l})<\varepsilon/4+\varepsilon/2+\varepsilon/4=\varepsilon$. So $a\in\mathcal{C}(X)$.

By clause 3, $\kappa_{X}(a_{m})\to[a]$. For every $m$, $\widehat{d}(\xi_{m},[a])\le\widehat{d}(\xi_{m},\kappa_{X}(a_{m}))+\widehat{d}(\kappa_{X}(a_{m}),[a])<r_{m}+\widehat{d}(\kappa_{X}(a_{m}),[a])$. Given $\varepsilon>0$, choose $M_{1}$ with $r_{m}<\varepsilon/2$ for $m\ge M_{1}$, then $M_{2}$ with $\widehat{d}(\kappa_{X}(a_{m}),[a])<\varepsilon/2$ for $m\ge M_{2}$, and let $M$ be the larger; then $\widehat{d}(\xi_{m},[a])<\varepsilon$ for $m\ge M$. Thus $(\xi_{m})$ converges to $[a]\in\widehat{X}$, and $(\widehat{X},\widehat{d})$ is complete by Complete Metric Space.

Proof of clause 5 (Extension). Step 1: limits along Cauchy sequences. For $x\in\mathcal{C}(X)$ the sequence $(f(x_{k}))_{k\in\mathbb{N}}$ is Cauchy in $(Y,e)$ by hypothesis, so it converges by Complete Metric Space, to a unique limit by (ii). Write $L(x)=\lim_{k\to\infty}f(x_{k})$.

Step 2: independence of the representative. Let $x,y\in\mathcal{C}(X)$ with $x\approx y$; we show $L(x)=L(y)$. Let $T=\{0,1\}$ and let $t:\mathbb{N}\times T\to T$ be given by $t(n,0)=1$ and $t(n,1)=0$. By Definition of Sequences by Recursion on the Natural Numbers §recursion there is a sequence $\tau:\mathbb{N}\to T$ with $\tau(1)=0$ and $\tau(n+1)=t(n,\tau(n))\ne\tau(n)$ for every $n$ (recall $S(n)=n+1$ in Natural Numbers). Hence for every $N\in\mathbb{N}$ one of $N$, $N+1$ is an index $m$ with $\tau(m)=0$ and the other an index with $\tau(m)=1$, and both are $\ge N$ by clauses 1 and 5 of Properties of the Order on the Natural Numbers. $(\dagger)$

Define the sequence $z$ in $X$ by $z_{m}=x_{m}$ if $\tau(m)=0$ and $z_{m}=y_{m}$ if $\tau(m)=1$. For every $m$ we have $d(z_{m},x_{m})\le d(x_{m},y_{m})$: if $\tau(m)=0$ the left side is $0$ (condition 2) and the right side is nonnegative (condition 1); if $\tau(m)=1$ the two sides are equal by condition 3. Hence, by conditions 3 and 4,

Since $\delta(x,y)=0$, the real sequence $(d(x_{m},y_{m}))_{m}$ converges to $0$ (Cauchy Sequences in a Metric Space: Constant Sequences, Convergence of Termwise Distances, the Null Relation, and Independence of Representatives §distances), and by clause 3 of Properties of the Absolute Value in an Ordered Field $d(x_{m},y_{m})\le|d(x_{m},y_{m})-0|$. Given $\varepsilon>0$, first choose $N_{1}$ with $d(x_{m},y_{m})<\varepsilon/4$ for $m\ge N_{1}$, then $N_{2}$ with $d(x_{m},x_{l})<\varepsilon/2$ for $m,l\ge N_{2}$, and let $N$ be the larger; for $m,l\ge N$ the display gives $d(z_{m},z_{l})<\varepsilon$. So $z\in\mathcal{C}(X)$ and $f(z_{m})\to L(z)$ by Step 1.

We show $L(x)=L(z)$. Let $\varepsilon>0$. First choose $K_{1}$ with $e(f(x_{k}),L(x))<\varepsilon/2$ for $k\ge K_{1}$, then $K_{2}$ with $e(f(z_{k}),L(z))<\varepsilon/2$ for $k\ge K_{2}$, let $K$ be the larger, and by $(\dagger)$ choose $m\ge K$ with $\tau(m)=0$, so that $z_{m}=x_{m}$. Then $e(L(x),L(z))\le e(L(x),f(x_{m}))+e(f(z_{m}),L(z))<\varepsilon$. As $\varepsilon>0$ was arbitrary and $0\le e(L(x),L(z))$, Comparison of Real Numbers with Arbitrary Positive Slack §vanishing gives $e(L(x),L(z))=0$, so $L(x)=L(z)$ by condition 2. The same argument with an index $m\ge K$ satisfying $\tau(m)=1$, for which $z_{m}=y_{m}$, gives $L(y)=L(z)$. Hence $L(x)=L(y)$.

Step 3: existence and uniqueness of $\widehat{f}$. Let $\widehat{f}=\{(\xi,L(x)):\ x\in\mathcal{C}(X),\ \xi=[x]\}\subseteq\widehat{X}\times Y$. Every $\xi\in\widehat{X}$ occurs as a first coordinate by (vi), and if $[x]=[y]$ then $x\approx y$ by Cauchy Sequences in a Metric Space: Constant Sequences, Convergence of Termwise Distances, the Null Relation, and Independence of Representatives §null, so $L(x)=L(y)$ by Step 2. Thus $\widehat{f}$ is a map $\widehat{X}\to Y$ (no choice is involved) and $\widehat{f}([x])=L(x)=\lim_{k}f(x_{k})$ for every $x\in\mathcal{C}(X)$. If $g:\widehat{X}\to Y$ also satisfies this formula, then for $\xi=[x]\in\widehat{X}$ we get $g(\xi)=L(x)=\widehat{f}(\xi)$; so $\widehat{f}$ is the only such map.

Step 4: $\widehat{f}\circ\kappa_{X}=f$. For $a\in X$, (vi) gives $\widehat{f}(\kappa_{X}(a))=\widehat{f}([\bar{a}])=L(\bar{a})$, the limit of the constant sequence with value $f(a)$, which is $f(a)$ by (iii) and (ii).

Step 5: continuity. By (vii) it suffices to show that $\xi_{m}\to\xi$ in $(\widehat{X},\widehat{d})$ implies $\widehat{f}(\xi_{m})\to\widehat{f}(\xi)$ in $(Y,e)$. For $m\in\mathbb{N}$ let

Each $A_{m}$ is nonempty: writing $\xi_{m}=[x]$ by (vi), clause 3 gives $\kappa_{X}(x_{k})\to\xi_{m}$ and Step 3 gives $f(x_{k})\to\widehat{f}(\xi_{m})$; choose $K_{1}$ with $\widehat{d}(\kappa_{X}(x_{k}),\xi_{m})<r_{m}$ for $k\ge K_{1}$, then $K_{2}$ with $e(f(x_{k}),\widehat{f}(\xi_{m}))<r_{m}$ for $k\ge K_{2}$, and let $k$ be the larger; then $x_{k}\in A_{m}$. By the axiom of countable choice there is a sequence $a=(a_{m})_{m\in\mathbb{N}}$ in $X$ with $a_{m}\in A_{m}$ for every $m$.

First, $\kappa_{X}(a_{m})\to\xi$: for every $m$, $\widehat{d}(\kappa_{X}(a_{m}),\xi)\le\widehat{d}(\kappa_{X}(a_{m}),\xi_{m})+\widehat{d}(\xi_{m},\xi)<r_{m}+\widehat{d}(\xi_{m},\xi)$; given $\varepsilon>0$ choose $M_{1}$ with $r_{m}<\varepsilon/2$ for $m\ge M_{1}$ by (v), then $M_{2}$ with $\widehat{d}(\xi_{m},\xi)<\varepsilon/2$ for $m\ge M_{2}$, and take $m$ at least the larger. Second, $a\in\mathcal{C}(X)$: by clause 2, $d(a_{m},a_{l})=\widehat{d}(\kappa_{X}(a_{m}),\kappa_{X}(a_{l}))\le\widehat{d}(\kappa_{X}(a_{m}),\xi)+\widehat{d}(\xi,\kappa_{X}(a_{l}))$; given $\varepsilon>0$ choose $M$ with $\widehat{d}(\kappa_{X}(a_{m}),\xi)<\varepsilon/2$ for $m\ge M$, so $d(a_{m},a_{l})<\varepsilon$ for $m,l\ge M$. Third, by clause 3, $\kappa_{X}(a_{m})\to[a]$, so $[a]=\xi$ by Uniqueness of Limits in a Metric Space; hence $f(a_{m})\to L(a)=\widehat{f}([a])=\widehat{f}(\xi)$ by Steps 1 and 3. Finally, for every $m$,

Given $\varepsilon>0$ choose $M_{1}$ with $r_{m}<\varepsilon/2$ for $m\ge M_{1}$, then $M_{2}$ with $e(f(a_{m}),\widehat{f}(\xi))<\varepsilon/2$ for $m\ge M_{2}$; for $m$ at least the larger, $e(\widehat{f}(\xi_{m}),\widehat{f}(\xi))<\varepsilon$. So $\widehat{f}(\xi_{m})\to\widehat{f}(\xi)$, and $\widehat{f}$ is continuous.

Step 6: the Lipschitz property. Suppose $f$ is Lipschitz with constant $\Lambda\ge0$. Let $\alpha,\beta\in\widehat{X}$ and write $\alpha=[x]$, $\beta=[y]$ by (vi). By Step 3, $f(x_{k})\to\widehat{f}(\alpha)$ and $f(y_{k})\to\widehat{f}(\beta)$, so by (iv) the real sequence $(e(f(x_{k}),f(y_{k})))_{k}$ converges to $e(\widehat{f}(\alpha),\widehat{f}(\beta))$. By Cauchy Sequences in a Metric Space: Constant Sequences, Convergence of Termwise Distances, the Null Relation, and Independence of Representatives §distances, $(d(x_{k},y_{k}))_{k}$ converges to $\delta(x,y)=\widehat{d}(\alpha,\beta)$, so $(\Lambda\,d(x_{k},y_{k}))_{k}$ converges to $\Lambda\,\widehat{d}(\alpha,\beta)$ by clause 3 of Arithmetic of Limits of Real Sequences. Since $e(f(x_{k}),f(y_{k}))\le\Lambda\,d(x_{k},y_{k})$ for every $k$, clause 1 (comparison) of Order Properties of Limits of Real Sequences gives $e(\widehat{f}(\alpha),\widehat{f}(\beta))\le\Lambda\,\widehat{d}(\alpha,\beta)$. Thus $\widehat{f}$ is Lipschitz with constant $\Lambda$.

Proof of clause 6 (Uniqueness of continuous extensions). Let $\xi\in\widehat{X}$ and write $\xi=[x]$ by (vi). By clause 3, $\kappa_{X}(x_{k})\to\xi$, so by (vii) $g(\kappa_{X}(x_{k}))\to g(\xi)$ and $h(\kappa_{X}(x_{k}))\to h(\xi)$ in $(Y,e)$. Since $g(\kappa_{X}(x_{k}))=h(\kappa_{X}(x_{k}))$ for every $k$, one sequence converges to both $g(\xi)$ and $h(\xi)$, and Uniqueness of Limits in a Metric Space gives $g(\xi)=h(\xi)$.

Proof of clause 7 (Inequalities). Let $\xi\in\widehat{X}$ and write $\xi=[x]$ by (vi). By clause 3, $\kappa_{X}(x_{k})\to\xi$, so by (vii), applied with $(Z,\rho)$ the real line of The Absolute Value Metric on the Real Line, $g(\kappa_{X}(x_{k}))\to g(\xi)$ and $h(\kappa_{X}(x_{k}))\to h(\xi)$ in the real line, hence in the sense of Limit of a Sequence of Real Numbers by (ii). The hypothesis with $a=x_{k}$ gives $g(\kappa_{X}(x_{k}))\le h(\kappa_{X}(x_{k}))$ for every $k$, so clause 1 (comparison) of Order Properties of Limits of Real Sequences gives $g(\xi)\le h(\xi)$. $\blacksquare$