False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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

Statement

False claim: if $A \subseteq R$ and $f : A \to R$ is continuous on $A$ (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ 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 $f : A \to R$ : one $δ$ serving every pair of points of $A$ ).

Why it is tempting. Continuity says that for every $ε > 0$ and every point $c$ there is a $δ > 0$ that works at $c$ . It is easy to read that as producing "a $δ$ ", forgetting that the $δ$ was produced after $c$ was fixed and may depend on it. Uniform continuity demands one $δ$ 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 $R$ : a continuous real function on a compact subset of $R$ is uniformly continuous, proved $R$ -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 $E \subseteq R$ there is a continuous function on $E$ that is not uniformly continuous, which is Rudin 4.20, the sharp converse: on a noncompact $E \subseteq R$ 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.

Facts & Assumptions

Given: The domain $A : = (0, 1) = {x \in R : 0 < x < 1}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) and the function $f : A \to R$ , $f (x) : = 1 / x$ . Natural numbers are identified with their canonical images in $R$ .

[L1]

Continuity on $A$ and uniform continuity on $A$ , in the forms of Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point and Uniform continuity of $f : A \to R$ : one $δ$ serving every pair of points of $A$ ; in particular, $f$ fails to be uniformly continuous on $A$ as soon as some real $ε_{0} > 0$ admits, for every real $δ > 0$ , a pair $x, x^{'} \in A$ with $∣ x - x^{'} ∣ < δ$ and $∣ f (x) - f (x^{'}) ∣ \geq ε_{0}$ .

[L2]

Algebra of continuous functions: the identity is continuous on $A$ , and if $q$ is continuous on $A$ and does not vanish there then $1 / q$ is continuous on $A$ (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function).

[L3]

Archimedean property in reciprocal form: for every real $η > 0$ there is a natural $k \geq 1$ with $1 / k < η$ ; and $0 < s < t$ implies $0 < 1 / t < 1 / s$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

[L4]

Ordered-field arithmetic in $R$ : for a natural $k$ , $0 < 1 / (k + 3) < 1 / (k + 2) < 1$ , so both lie in $(0, 1)$ ; the product $(k + 2) (k + 3) \geq k + 2 \geq 2$ ; and $∣ u ∣ \geq 0$ with $∣ u ∣ = u$ for $u \geq 0$ (Ordered field, Basic properties of the absolute value, Integer powers $a^{m}$ ).

Refutation

technique · direct

1.1

$f$ is continuous on $A$ . The identity is continuous on $A$ by [L2] and does not vanish on $A$ , since $x > 0$ for $x \in (0, 1)$ ; so $f (x) = 1 / x$ is continuous on $A$ by [L2].

L2L4

1.2

For $k \in N$ put $x_{k} : = 1 / (k + 2)$ and $y_{k} : = 1 / (k + 3)$ . By [L4] both lie in $(0, 1) = A$ , and $f (x_{k}) = k + 2$ , $f (y_{k}) = k + 3$ . Note that $N$ contains $0$ , so the smallest pair is $x_{0} = 1 / 2$ and $y_{0} = 1 / 3$ , and no index $k$ produces a point outside $A$ .

2.1

The gap between the arguments is $∣ x_{k} - y_{k} ∣ = 1 / (k + 2) - 1 / (k + 3) = 1 / ((k + 2) (k + 3)) \leq 1 / (k + 2)$ , using $(k + 2) (k + 3) \geq k + 2 > 0$ and [L3]. The gap between the values is $∣ f (x_{k}) - f (y_{k}) ∣ = ∣ (k + 2) - (k + 3) ∣ = 1$ .

step 1.2L3L4

3.1

Take $ε_{0} : = 1$ and let a real $δ > 0$ be given. By [L3] there is a natural $m \geq 1$ with $1 / m < δ$ ; put $k : = m$ , so $k + 2 > m > 0$ and hence $1 / (k + 2) < 1 / m < δ$ by [L3]. Then $∣ x_{k} - y_{k} ∣ \leq 1 / (k + 2) < δ$ while $∣ f (x_{k}) - f (y_{k}) ∣ = 1 \geq ε_{0}$ .

step 2.1L3choose

4.1

So no $δ > 0$ serves $ε_{0} = 1$ : by [L1] the function $f$ is continuous on $A$ and not uniformly continuous on $A$ , and the claim is false.

step 1.1step 3.1L1∎

Remarks

The domain is bounded, and that is the point. $(0, 1)$ is bounded but not closed, hence not compact (A subset of $R$ is compact if and only if it is closed and bounded), so Heine-Cantor in $R$ : a continuous real function on a compact subset of $R$ is uniformly continuous, proved $R$ -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 $R$ and not uniformly continuous there, the pairs $k + 1$ and $k + 1 + 1 / (k + 1)$ defeating every $δ$ ; that witness is $x \mapsto x^{2}$ is continuous on $R$ and not uniformly continuous, the pairs $k + 1$ and $k + 1 + 1 / (k + 1)$ defeating every $δ$ ↗ 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 $δ$ ↗ repeats the computation above with the estimates spelled out and records what it witnesses about the regularity hierarchy of Dictionary: for $A \subseteq R$ with the metric $d (x, y) = ∣ x - y ∣$ , continuity and uniform continuity of $f : A \to R$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $R$ is compact in the open-cover sense of $R$ exactly when it is a compact metric subspace.

Depends on

Used by

Dependency tree · two levels

47 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Uniform continuity (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 4 (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.4 (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.3: Uniform continuity (standard reference, not scraped)