CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31

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Dini's theorem fails for discontinuous approximants: shrinking interval indicators decrease pointwise to zero but not uniformly

Statement refuted

Refuted claim: continuity of the approximating functions in Dini's theorem can be dropped.

For $k \in N$ define $h_{k} : [0, 1] \to R$ to be the indicator of

$(0, 1 / ι (k + 1)) .$

Thus $h_{k}$ has value $0$ at both endpoints of that open interval. The sequence decreases pointwise to the continuous zero function but does not converge uniformly.

Facts & Assumptions

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

[L1]

For every real $ε > 0$ there is $N \geq 1$ with $1 / ι (N) < ε$ , and 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).

[L2]

Continuity at $c$ requires that every positive output error admit a positive input radius on which all function values remain close to the value at $c$ (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).

[L3]

Dini's theorem on a closed interval assumes that every approximating function and the pointwise limit are continuous (Dini's theorem on a closed interval: monotone pointwise convergence of continuous functions to a continuous limit is uniform).

Counterexample

technique · direct

1.1

The intervals $(0, 1 / a_{k + 1})$ are contained in $(0, 1 / a_{k})$ , so $h_{k + 1} (x) \leq h_{k} (x)$ for every $x \in [0, 1]$ .

1.2

At $x = 0$ , every $h_{k} (x)$ is $0$ . If $x > 0$ , choose $N$ with $1 / ι (N) < x$ ; then $h_{k} (x) = 0$ for all $k \geq N$ . Thus $h_{k} \to 0$ pointwise.

L1choose

1.3

Each $h_{k}$ is discontinuous at $0$ : for any $δ > 0$ , the point $y : = \min {δ / 2, 1 / (2 a_{k})}$ satisfies $0 < y < δ$ , lies in $(0, 1 / a_{k})$ , and has $∣ h_{k} (y) - h_{k} (0) ∣ = 1$ .

L2algebra

1.4

For each $k$ , the point $x_{k} : = 1 / (2 a_{k})$ lies in $(0, 1 / a_{k})$ and satisfies $h_{k} (x_{k}) = 1$ , so the convergence to $0$ is not uniform.

givenL1

2.1

The compact domain, monotone pointwise convergence, and continuous limit remain, but the approximants are discontinuous and uniform convergence fails; their continuity is indispensable in [L3].

step 1.1step 1.2step 1.3step 1.4L3∎

Depends on

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Sources

J. Lebl, Basic Analysis I, §6.1 (standard reference, not scraped)
Dini's theorem (Wikipedia) (standard reference, not scraped)
W. Trench, Introduction to Real Analysis (standard reference, not scraped)