$x\mapsto 1\mathrm{/}x$ is continuous on $(0,1)$ and not uniformly continuous there, the pairs $1\mathrm{/}(k+2)$ and $1\mathrm{/}(k+3)$ defeating every $\delta$

Refuted claim: the function $f:(0,1)\to \mathbb{R}$, $f(x):=1\mathrm{/}x$, is uniformly continuous on $(0,1)$ (Uniform continuity of $f:A\to \mathbb{R}$: one $\delta$ serving every pair of points of $A$, Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length).

$f$ is continuous on $(0,1)$ (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) and its domain is bounded, so this is the sharpest simple instance of FALSE: every continuous real function is uniformly continuous on its domain: neither continuity nor boundedness of the domain implies uniform continuity, and what Heine-Cantor in $\mathbb{R}$: a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$-natively from sequential compactness actually needs is compactness, which $(0,1)$ does not have because it is not closed (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded).

The refutation exhibits, for every $\delta >0$, a pair of points of $(0,1)$ closer than $\delta$ whose $f$-values differ by exactly $1$. The pairs are

$x_k:=\frac{1}{k+2},y_k:=\frac{1}{k+3}(k\in \mathbb{N}),$

and the shift by $2$ and $3$ is not cosmetic: $\mathbb{N}$ contains $0$ here (Sequences of reals: bounded, eventually, frequently, tails, subsequences is $0$-indexed), so $1\mathrm{/}k$ is undefined at $k=0$ and $1\mathrm{/}(k+1)$ leaves $(0,1)$ at $k=0$.

• What this witnesses in the regularity hierarchy. 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, transported to real functions by 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, gives uniformly continuous $\Rightarrow$ continuous and asserts no converse. This item is the witness that the converse fails, and it is one of the two named in the remarks of 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; the other, On $[0,1]$ the function $x^{\beta }$ is $\beta$-Hölder and is $\alpha$-Hölder for no rational $\alpha >\beta$, so the Hölder classes are strictly nested, separates the Hölder classes below it.

• The failure is at the missing endpoint, and it is repaired by restoring it. On $[a,1]$ with $0<a<1$ the same formula is uniformly continuous, by Heine-Cantor in $\mathbb{R}$: a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$-natively from sequential compactness, since $[a,1]$ is closed and bounded. It is also repaired by an explicit estimate: on $[a,1]$ one has $\mathrm{\mid }1\mathrm{/}x-1\mathrm{/}{x}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}(x{x}^{\mathrm{\prime }})\le \mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}{a}^2$, so $f$ is even Lipschitz there.

• A second reading of the same fact. Assuming the Axiom of Choice (The Axiom of Choice), A uniformly continuous real function on a subset $D\subseteq \mathbb{R}$ extends uniquely to a uniformly continuous function on the closure of $D$ says that a uniformly continuous function on $(0,1)$ would extend continuously to $[0,1]$ and hence be bounded there (A continuous real function on a compact subset of $\mathbb{R}$ is bounded); $1\mathrm{/}x$ is unbounded on $(0,1)$, so it cannot be uniformly continuous. That route is shorter but rests on more, and the computation above is the elementary one.