Completeness, Completion, and Uniform Continuity: Examples and Counterexamples

Regard $\mathbb{Q}$ (The rationals as equivalence classes of pairs of integers) as the subset $\widehat{\mathbb{Q}}$ of $\mathbb{R}$ that is the image of the canonical embedding $q\mapsto \widehat{q}$ (The rationals embed densely in the reals), carrying the subspace metric $d_{\mathbb{Q}}(p,q)=\mathrm{\mid }\widehat{p}-\widehat{q}\mathrm{\mid }$ inherited from the usual metric of $\mathbb{R}$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset). Let $\iota :\mathbb{Q}\to \mathbb{R}$ be that embedding.

Then $((\mathbb{R},d_{\mathbb{R}}),\iota )$ is a completion of $(\mathbb{Q},d_{\mathbb{Q}})$ (A completion of a metric space: a complete metric space together with an isometric embedding onto a dense subspace). Consequently, by uniqueness of completions (A completion is unique up to a unique isometry fixing the original space, and uniformly continuous maps into complete spaces extend through it), for every completion $((\widehat{Y},\widehat{d}),j)$ of $(\mathbb{Q},d_{\mathbb{Q}})$ there is exactly one continuous $\phi :\mathbb{R}\to \widehat{Y}$ with $\phi \circ \iota =j$, and it is an isometry. In this sense $\mathbb{R}$ is the completion of $\mathbb{Q}$.

• This is the metric statement of what the construction pages did by hand. This library builds $\mathbb{R}$ out of Cauchy sequences of rationals on its own page, and the general construction of Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences follows the same plan for an arbitrary metric space: classes of Cauchy sequences, with the distance read off as a limit. The present item is the observation that, run on $\mathbb{Q}$, that plan reaches a space isometric to the $\mathbb{R}$ already in hand, and that no separate verification of "which complete space it is" is needed once uniqueness is available.
• Density is the whole of the third condition, and it is exactly the Archimedean fact. That every interval of positive length contains a rational is The rationals embed densely in the reals; without it $\mathbb{Q}$ would sit inside $\mathbb{R}$ as a complete-looking but small subspace, and $\mathbb{R}$ would not be a completion of it but merely a complete space containing it.
• The metric is the one written above and no other. Completions are taken with respect to a named metric (A completion of a metric space: a complete metric space together with an isometric embedding onto a dense subspace), and a different metric on $\mathbb{Q}$ would in general have a different completion; nothing here is a statement about $\mathbb{Q}$ as a bare field or as a bare topological space.

Let $X:=[1,2]\subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) with the metric $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ inherited from $\mathbb{R}$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset), and define

$f(x):=\frac{x+2\mathrm{/}x}{2}=\frac{x}{2}+\frac{1}{x}\mathrm{.}$

Then:

1. $(X,d)$ is a nonempty complete metric space (Complete metric space: every Cauchy sequence converges in the space).
2. $f$ maps $X$ into $X$.
3. $f$ is a contraction with constant $q=1\mathrm{/}2$ (Lipschitz map, $\alpha$-Hölder map for rational $0<\alpha \le 1$, and contraction).
4. $f$ has exactly one fixed point in $X$, namely $\sqrt{2}$ (Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$), and the iterates $x_{n+1}=f(x_n)$ converge to it from every starting point $x_0\in X$ (A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point).
5. Starting from $x_0=1$ one has $x_1=3\mathrm{/}2$ and $d(x_1,x_0)=1\mathrm{/}2$, so the a priori bound of The a priori bound $d(x^{\ast },x_n)\le q^{n}d(x_1,x_0)\mathrm{/}(1-q)$ and the a posteriori bound $d(x^{\ast },x_{n+1})\le qd(x_{n+1},x_n)\mathrm{/}(1-q)$ reads $\mid \sqrt{2}-x_n\mid \le \frac{(1\mathrm{/}2{)}^n\cdot (1\mathrm{/}2)}{1-1\mathrm{/}2}=(\frac{1}{2}{)}^n,$ and the a posteriori bound reads $\mid \sqrt{2}-x_{n+1}\mid \le \mathrm{\mid }{x}_{n+1}-x_n\mathrm{\mid }$.

• No derivative is used anywhere. The contraction constant comes from the algebraic identity of step 1.4, $f(x)-f(y)=(x-y)(1\mathrm{/}2-1\mathrm{/}(xy))$, and from the bound $1\le xy\le 4$ that the interval supplies. This matters here: the mean value theorem, which is how this estimate is usually obtained, belongs to a later page of this library and is not available at this point.
• The lower endpoint is what makes the estimate close. The bound $\mathrm{\mid }1\mathrm{/}2-1\mathrm{/}(xy)\mathrm{\mid }\le 1\mathrm{/}2$ of step 2.2 needs $xy\ge 1$, and that is exactly what $x,y\ge 1$ supplies; on a set of small positive reals the factor $1\mathrm{/}(xy)$ is large and the same computation gives nothing. The self-mapping property uses both endpoints: step 1.3 gives the lower bound $\sqrt{2}$ and step 2.1 the upper bound $2$.
• What the bounds say numerically. From $x_0=1$ the a priori bound guarantees $\mathrm{\mid }\sqrt{2}-x_n\mathrm{\mid }\le 2^{-n}$, so ten steps give an error at most $2^{-10}$. This is a guarantee, not the truth: the iteration is Newton's method for $x^2-2$, and it converges far faster than the bound admits. The a posteriori bound of claim 5, which costs one subtraction, is what a computation would actually report.
• The restriction to $[1,2]$ is convenience, not necessity. On $[1,\mathrm{\infty })$ one still has $x,y\ge 1$, hence $0<1\mathrm{/}(xy)\le 1$ and $\mathrm{\mid }1\mathrm{/}2-1\mathrm{/}(xy)\mathrm{\mid }\le 1\mathrm{/}2$, so the same computation makes $f$ a contraction with the same constant there, and step 1.3 already shows $f$ maps $[1,\mathrm{\infty })$ into itself. The shorter interval is used only to keep the arithmetic of step 2.2 explicit.

Regard $\mathbb{Q}$ as a subspace of $\mathbb{R}$ with the metric inherited from the usual metric (The rationals embed densely in the reals, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset), let $L\in \mathbb{R}$ with $L\ge 0$, and let $f:\mathbb{Q}\to \mathbb{R}$ be Lipschitz with constant $L$ (Lipschitz map, $\alpha$-Hölder map for rational $0<\alpha \le 1$, and contraction), that is

$\mathrm{\mid }f(p)-f(q)\mathrm{\mid }\le L\mathrm{\mid }p-q\mathrm{\mid }\text{for all }p,q\in \mathbb{Q}\mathrm{.}$

Then there is exactly one continuous $F:\mathbb{R}\to \mathbb{R}$ with $F(q)=f(q)$ for every $q\in \mathbb{Q}$, and that $F$ is again Lipschitz with the same constant $L$.

• The extension theorem gives uniform continuity; the constant is recovered separately. A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space transports a modulus of continuity, not a Lipschitz constant, so step 4.1 is genuinely needed. The mechanism is generic: a non-strict inequality that holds on a dense set and whose two sides are continuous holds everywhere, by Limits preserve non-strict inequalities.
• Nothing here is special to $\mathbb{Q}$ and $\mathbb{R}$. The same argument extends a Lipschitz map from any dense subspace of any metric space into any complete metric space, with the same constant; $\mathbb{Q}\subseteq \mathbb{R}$ is used only because it is the density statement this library already has (The rationals embed densely in the reals).
• Uniqueness is what makes "the" extension meaningful, and it needs only continuity, not the Lipschitz condition: two continuous maps agreeing on a dense set agree everywhere (A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed, For a map of metric spaces the following agree: $\epsilon$-$\delta$ continuity everywhere, preimages of open sets are open, preimages of closed sets are closed, sequential continuity, and $f(\bar{A})\subseteq \overline{f(A)}$).
• The hypothesis cannot be relaxed to continuity. A continuous $f:\mathbb{Q}\to \mathbb{R}$ need not extend continuously to $\mathbb{R}$ at all, and the reason is the same one that defeats every argument of this kind: continuity does not control a function along a Cauchy sequence (A uniformly continuous map sends Cauchy sequences to Cauchy sequences, $x\mapsto 1\mathrm{/}x$ is continuous on $(0,1)$ and sends the Cauchy sequence $(1\mathrm{/}(k+2){)}_{k\ge 0}$ to an unbounded one).

Let $X:=[0,\mathrm{\infty })\subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) with the metric $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ inherited from $\mathbb{R}$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset), and let $g:X\to X$ be $g(x):=\sqrt{x}=x^{1\mathrm{/}2}$ (Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$, Rational powers $a^r$ of a positive base). Then:

1. $\mid \sqrt{x}-\sqrt{y}\mid \le \mathrm{\mid }x-y{\mathrm{\mid }}^{1\mathrm{/}2}$ for all $x,y\ge 0$, so $g$ is $1\mathrm{/}2$-Hölder with constant $1$ (Lipschitz map, $\alpha$-Hölder map for rational $0<\alpha \le 1$, and contraction).
2. $g$ is uniformly continuous (Uniform continuity of a map of metric spaces: one $\delta$ serving every point).
3. The constant $1$ cannot be improved: every $1\mathrm{/}2$-Hölder constant $C$ for $g$ satisfies $C\ge 1$.
4. For every rational $\alpha$ with $1\mathrm{/}2<\alpha \le 1$, the map $g$ is not $\alpha$-Hölder. In particular, at $\alpha =1$, $g$ is not Lipschitz.

So $1\mathrm{/}2$ is exactly the Hölder exponent of the square root, and the example separates "Hölder" from "Lipschitz" inside Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent.

• Where the failure is located. The obstruction in claims 3 and 4 sits at $0$: the inequality $t^{1\mathrm{/}2}\le C{t}^{\alpha }$ is comfortable for large $t$ and impossible for small $t$ once $\alpha >1\mathrm{/}2$, because $t^{1\mathrm{/}2}$ then dominates $t^{\alpha }$. Away from $0$ the square root is perfectly Lipschitz: on $[a,\mathrm{\infty })$ with $a>0$ one has $\mid \sqrt{x}-\sqrt{y}\mid =\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}(\sqrt{x}+\sqrt{y})\le \mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}(2\sqrt{a})$, since $\sqrt{x}+\sqrt{y}\ge 2\sqrt{a}>0$ and $(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})=x-y$.
• The exponent is rational, and that is not a restriction here. At this page's position in the reading order, Rational powers $a^r$ of a positive base is the available exponent construction. Both $1\mathrm{/}2$ and every exponent used above are rational, so claim 4 is intentionally local to rational $\alpha$; real powers are introduced later in Real powers for positive bases, with the zero-base positive-exponent convention ↗.
• What this example is for. It is one of the two witnesses named in Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent: it shows that "Hölder" is strictly weaker than "Lipschitz", so the implication proved there from Hölder to uniform continuity is not a detour through the Lipschitz condition. The other witness, separating continuity from uniform continuity, is $x\mapsto 1\mathrm{/}x$ is continuous on $(0,1)$ and sends the Cauchy sequence $(1\mathrm{/}(k+2){)}_{k\ge 0}$ to an unbounded one.
• Claim 1 is sharp in a second sense as well: equality holds whenever one of the two arguments is $0$, since $\mid \sqrt{x}-\sqrt{0}\mid =\sqrt{x}=\mathrm{\mid }x-0{\mathrm{\mid }}^{1\mathrm{/}2}$. So the estimate is attained and not merely approached.

Let $S$ be a nonempty set, let $\mathcal{B}(S):=\{f:f\text{ is a bounded function }S\to \mathbb{R}\}$, and let $d_{\mathrm{\infty }}(f,g):=\sup \{\mathrm{\mid }f(s)-g(s)\mathrm{\mid }:s\in S\}$ be the supremum metric, which is a metric on $\mathcal{B}(S)$ (The supremum metric $d_{\mathrm{\infty }}(f,g)={\sup }_x\mathrm{\mid }f(x)-g(x)\mathrm{\mid }$ is a metric on the bounded real-valued functions on a nonempty set, Lower bound, bounded below, bounded set).

Then $(\mathcal{B}(S),d_{\mathrm{\infty }})$ is a complete metric space (Complete metric space: every Cauchy sequence converges in the space).

The limit is produced pointwise and then shown to be bounded and to be approached uniformly; that order is the content of the proof.

• The two limits are taken in different orders, and that is the point. Step 2.1 fixes $s$ and lets $k$ grow, producing a candidate limit function; steps 2.2 to 4.1 fix $k$ and let the other index grow inside an estimate that is uniform in $s$. It is the uniformity of the bound $\epsilon \mathrm{/}2$ in $s$, and nothing else, that converts pointwise convergence into convergence in $d_{\mathrm{\infty }}$.
• Boundedness of the limit is a separate step and is genuinely needed. The metric $d_{\mathrm{\infty }}(f_k,f)$ is only defined once $f$ is known to be bounded (The supremum metric $d_{\mathrm{\infty }}(f,g)={\sup }_x\mathrm{\mid }f(x)-g(x)\mathrm{\mid }$ is a metric on the bounded real-valued functions on a nonempty set), so step 3.1 has to come before step 4.1. The extended real line is introduced later, but it is not used as a metric codomain or as a placeholder here; $d_{\mathrm{\infty }}(f_k,f)$ is real-valued only after boundedness of $f$ is proved.
• Nothing is assumed about $S$ beyond nonemptiness, which is what The supremum metric $d_{\mathrm{\infty }}(f,g)={\sup }_x\mathrm{\mid }f(x)-g(x)\mathrm{\mid }$ is a metric on the bounded real-valued functions on a nonempty set needs so that the supremum is taken over a nonempty set. In particular $S$ carries no topology and no metric here; the functions are arbitrary bounded functions, not continuous ones.
• Where the least-upper-bound property is spent. Twice: inside The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges at step 2.1, and in the very definition of $d_{\mathrm{\infty }}$ (The supremum metric $d_{\mathrm{\infty }}(f,g)={\sup }_x\mathrm{\mid }f(x)-g(x)\mathrm{\mid }$ is a metric on the bounded real-valued functions on a nonempty set). Completeness of $\mathbb{R}$ is the whole engine of this example.

Refuted claim: the completeness hypothesis in Banach's fixed point theorem (A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point) may be dropped; a contraction of a nonempty metric space into itself always has a fixed point.

Let $X:=(0,1]\subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) carry the metric $d(x,y):=\mathrm{\mid }x-y\mathrm{\mid }$ inherited from the real line (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset), and let $f:X\to X$ be $f(x):=x\mathrm{/}2$. Then $X$ is nonempty, $f$ maps $X$ into itself and is a contraction with constant $1\mathrm{/}2$ (Lipschitz map, $\alpha$-Hölder map for rational $0<\alpha \le 1$, and contraction), and $f$ has no fixed point in $X$. The single hypothesis of A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point that fails is completeness (Complete metric space: every Cauchy sequence converges in the space), and it does fail.