A uniformly continuous real function on a subset $D\subseteq \mathbb{R}$ extends uniquely to a uniformly continuous function on the closure of $D$

Assume the Axiom of Choice (The Axiom of Choice).

Let $D\subseteq \mathbb{R}$ be nonempty and let $f:D\to \mathbb{R}$ be uniformly continuous on $D$ (Uniform continuity of $f:A\to \mathbb{R}$: one $\delta$ serving every pair of points of $A$). Write $\bar{D}$ for the closure of $D$ in $\mathbb{R}$ (Interior, closure, boundary and exterior of a subset of $\mathbb{R}$). Then:

1. there is a uniformly continuous $g:\bar{D}\to \mathbb{R}$ with $g(x)=f(x)$ for every $x\in D$;
2. $g$ is the only continuous function $\bar{D}\to \mathbb{R}$ extending $f$ (Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point).

Uniform continuity is sufficient; continuity alone is not enough. The function $x\mapsto 1\mathrm{/}x$ is continuous on $D=(0,1)$, whose closure is $[0,1]$, and no continuous $g:[0,1]\to \mathbb{R}$ extends it. Indeed continuity at $0$ would give $\mathrm{\mid }g(x)-g(0)\mathrm{\mid }<1$ for all sufficiently small positive $x$, whereas taking also $x<1\mathrm{/}(\mathrm{\mid }g(0)\mathrm{\mid }+2)$ gives $1\mathrm{/}x>\mathrm{\mid }g(0)\mathrm{\mid }+2$. By this corollary, $x\mapsto 1\mathrm{/}x$ is therefore not uniformly continuous on $(0,1)$.

This is the metric extension theorem, read through the dictionary. The work is done by A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space, applied to the metric space $X:=\bar{D}$ with the subspace metric, its dense subset $D$, and the complete target $(\mathbb{R},d_{\mathbb{R}})$ ($\mathbb{R}$ and ${\mathbb{R}}^n$ for $n\ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$); Dictionary: for $A\subseteq \mathbb{R}$ with the metric $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$, continuity and uniform continuity of $f:A\to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace translates the hypothesis and the conclusion between the two vocabularies. The extension value is uniquely defined there. Nevertheless its proof uses countable choice, supplied here by AC, both to obtain the Cantor-intersection point and to extract an approximating sequence in the uniqueness argument.

Why later pages need exactly this. The exponential and the power functions are defined on $\mathbb{Q}$ first and then extended to $\mathbb{R}$, and the extension step is this corollary with $D$ the rationals of an interval; that is the use for which it is stated here rather than inside an example.

• Uniqueness needs only continuity, and it needs density. Two continuous functions on $\bar{D}$ agreeing on $D$ agree everywhere, because $D$ is dense; that is the uniqueness half of A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space and it is why claim 2 quantifies over continuous extensions rather than over uniformly continuous ones. On a set where $D$ is not dense the conclusion is simply false: any values may be assigned off $\bar{D}$.

• The extension is uniformly continuous, not merely continuous, and with the same modulus in the following sense: any $\delta$ that works for $f$ on $D$ and a given $\epsilon$ works for $g$ on $\bar{D}$ and any ${\epsilon }^{\mathrm{\prime }}>\epsilon$. That refinement is not asserted here; what is asserted is what A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space proves.

• The same conclusion, reached directly. That $x\mapsto 1\mathrm{/}x$ is not uniformly continuous on $(0,1)$ is proved on the companion page by exhibiting the pairs of points that defeat every $\delta$ ($x\mapsto 1\mathrm{/}x$ is continuous on $(0,1)$ and not uniformly continuous there, the pairs $1\mathrm{/}(k+2)$ and $1\mathrm{/}(k+3)$ defeating every $\delta$ ↗); that item is named here for orientation only, and nothing in this corollary rests on it.

• A special case worth naming. If $D$ is already closed then $\bar{D}=D$ (The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points) and the corollary says nothing. Its content is entirely about the points of $\bar{D}\setminus D$, which is where the values have to be created.