CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $k\in\mathbb{N}$ define $h_k:[0,1]\to\mathbb{R}$ to be the indicator of

$(0,1/\iota(k+1)).$

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 $a_k:=\iota(k+1)>0$ .

[L1]

For every real $\varepsilon>0$ there is $N\ge1$ with $1/\iota(N)<\varepsilon$ , and positive canonical naturals increase while their reciprocals decrease (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , The canonical natural $\iota(n) = n \cdot 1_F$ 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 $f : A \to \mathbb{R}$ 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).

[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 $h_{k+1}(x)\le h_k(x)$ for every $x\in[0,1]$ .

1.2

At $x=0$ , every $h_k(x)$ is $0$ . If $x>0$ , choose $N$ with $1/\iota(N)<x$ ; then $h_k(x)=0$ for all $k\ge N$ . Thus $h_k\to0$ pointwise.

L1choose

1.3

Each $h_k$ is discontinuous at $0$ : for any $\delta>0$ , the point $y:=\min\{\delta/2,1/(2a_k)\}$ satisfies $0<y<\delta$ , lies in $(0,1/a_k)$ , and has $|h_k(y)-h_k(0)|=1$ .

L2algebra

1.4

For each $k$ , the point $x_k:=1/(2a_k)$ 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

Used by

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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)