FALSE: every continuous real function is uniformly continuous on its domain

False claim: if $A \subseteq \mathbb{R}$ and $f : A \to \mathbb{R}$ is continuous on $A$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point), then $f$ is uniformly continuous on $A$ (Uniform continuity of $f : A \to \mathbb{R}$: one $\delta$ serving every pair of points of $A$).

Why it is tempting. Continuity says that for every $\varepsilon > 0$ and every point $c$ there is a $\delta > 0$ that works at $c$. It is easy to read that as producing "a $\delta$", forgetting that the $\delta$ was produced after $c$ was fixed and may depend on it. Uniform continuity demands one $\delta$ before any point is named, and the two quantifier orders are genuinely different.

What is true. On a compact domain the implication does hold, and that is 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; the metric-space form is Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous. Compactness is not a convenience there: for every noncompact bounded $E \subseteq \mathbb{R}$ there is a continuous function on $E$ that is not uniformly continuous, which is Rudin 4.20, the sharp converse: on a noncompact $E \subseteq \mathbb{R}$ there is an unbounded continuous function and a bounded continuous function with no greatest value, and if $E$ is bounded there is a continuous function on $E$ that is not uniformly continuous. The witness below is the smallest familiar instance of that theorem.

• The domain is bounded, and that is the point. $(0,1)$ is bounded but not closed, hence not compact (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded), so 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 does not apply. The obstruction sits at the missing endpoint $0$: the pairs above crowd towards it, their separation shrinking while the values they take diverge.

• Unboundedness produces the same failure for a different reason. $x \mapsto x^{2}$ is continuous on the closed set $\mathbb{R}$ and not uniformly continuous there, the pairs $k+1$ and $k+1+1/(k+1)$ defeating every $\delta$; that witness is $x \mapsto x^2$ is continuous on $\mathbb{R}$ and not uniformly continuous, the pairs $k+1$ and $k+1+1/(k+1)$ defeating every $\delta$ ↗ on the companion page, and it shows that closedness alone is no more sufficient than boundedness alone.

• The witness is worked out in full on the companion page. $x \mapsto 1/x$ is continuous on $(0,1)$ and not uniformly continuous there, the pairs $1/(k+2)$ and $1/(k+3)$ defeating every $\delta$ ↗ repeats the computation above with the estimates spelled out and records what it witnesses about the regularity hierarchy of Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$, 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.