CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

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Dini's theorem fails on $[0, \infty)$ : $x / (ι (k + 1) + x)$ decreases pointwise to zero but not uniformly

Statement refuted

Refuted claim: the compact-domain hypothesis in Dini's theorem can be dropped.

On $[0, \infty)$ define

$f_{k} (x) : = \frac{x}{ι (k + 1) + x} .$

The functions $f_{k}$ and their pointwise limit $0$ are continuous, and $f_{k + 1} (x) \leq f_{k} (x)$ for every $x \geq 0$ , but $f_{k} \to 0$ is not uniform.

Facts & Assumptions

Given: The functions $f_{k}$ in the Statement, with $a_{k} : = ι (k + 1) > 0$ .

[L1]

Constants and the identity are continuous; sums and quotients with nonvanishing denominator preserve continuity (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, 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).

[L2]

For every real $ε > 0$ there is $N \geq 1$ with $1 / ι (N) < ε$ , and the positive canonical naturals increase while their reciprocals decrease (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L3]

A subset of $R$ is compact exactly when it is closed and bounded; $[0, \infty)$ is unbounded (A subset of $R$ is compact if and only if it is closed and bounded, Lower bound, bounded below, bounded set).

[L4]

Dini's theorem on a closed interval concludes uniform convergence from continuity, pointwise monotonicity, and a continuous pointwise limit (Dini's theorem on a closed interval: monotone pointwise convergence of continuous functions to a continuous limit is uniform).

Counterexample

technique · direct

1.1

For every $k$ , the denominator $a_{k} + x$ is positive on $[0, \infty)$ , so $f_{k}$ is continuous by [L1]; the zero function is continuous as well.

givenL1

1.2

Since $a_{k + 1} > a_{k} > 0$ , one has $a_{k + 1} + x > a_{k} + x > 0$ , hence $f_{k + 1} (x) \leq f_{k} (x)$ for every $x \geq 0$ .

L2algebra

1.3

Fix $x \geq 0$ . If $x = 0$ then $f_{k} (x) = 0$ ; if $x > 0$ , then $0 \leq f_{k} (x) \leq x / a_{k}$ , and [L2] gives $x / a_{k} \to 0$ . Thus $f_{k} (x) \to 0$ for every $x$ .

L2algebra

1.4

At $x_{k} : = a_{k}$ one has $f_{k} (x_{k}) = a_{k} / (a_{k} + a_{k}) = 1 / 2$ , so the convergence is not uniform.

givenalgebra

1.5

The domain $[0, \infty)$ is not compact by [L3].

2.1

Hence all the listed Dini hypotheses except compactness hold, while the uniform conclusion fails; compactness cannot be dropped.

step 1.1step 1.2step 1.3step 1.4step 1.5L4∎

Depends on

Used by

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Sources

Dini's theorem (Wikipedia) (standard reference, not scraped)
W. Trench, Introduction to Real Analysis (standard reference, not scraped)

Dini's theorem fails on [0,∞)[0,∞): x/(ι(k+1)+x)x/(ι(k+1)+x) decreases pointwise to zero but not uniformly