ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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Dini's theorem applied to a nondecreasing sequence of piecewise linear approximations on $[0,1]$ , and what fails when the limit is not continuous

Example

Let $I := [0,1]$ with the metric inherited from $\mathbb{R}$ and let $a_k := 1/\iota(k+2)$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field), so $0 < a_k \le 1/2$ and $1/2 \le 1 - a_k < 1$ . Define the clipped identities

$s_k(t) := t \ \ (0 \le t \le 1 - a_k), \qquad s_k(t) := 1 - a_k \ \ (1 - a_k \le t \le 1) .$

Then:

each $s_k$ is continuous on the compact space $I$ , and the sequence is nondecreasing at every point: $s_k(t) \le s_{k+1}(t)$ ;
$s_k(t) \to t$ for every $t \in I$ , and the limit function $\mathrm{id}_I : t \mapsto t$ is continuous;
therefore Dini's theorem (Dini's theorem: on a compact metric space a nondecreasing sequence of continuous real functions converging pointwise to a continuous limit converges uniformly) applies and $(s_k)$ converges to $\mathrm{id}_I$ uniformly — which is confirmed by the direct estimate $|s_k(t) - t| \le a_k$ for every $t$ .

And the hypothesis that the limit is continuous cannot be dropped. The ramps $r_k$ of Refuted: $C(X,Y)$ is closed in the topology of pointwise convergence. The ramps on $[0,1]$ converge pointwise to a discontinuous limit are continuous on the same compact $I$ and pointwise nonincreasing, and they converge pointwise to the indicator of $\{1\}$ , which is not continuous; the conclusion of Dini's theorem fails for them, since a uniform limit of continuous functions would be continuous. So on a compact domain, with monotonicity and with continuity of every term, continuity of the limit is exactly the missing hypothesis, and it is not implied by the others.

Facts & Assumptions

Given: $I = [0,1]$ with $d(s,t) = |s-t|$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length), the reals $a_k = 1/\iota(k+2)$ , the functions $s_k$ displayed above, and the identity $\mathrm{id}_I$ of $I$ .

[L1]

$\iota$ is strictly increasing on $\mathbb{N}$ with $\iota(n) > 0$ for $n \ge 1$ , and $0 < u \le v$ gives $0 < 1/v \le 1/u$ ; hence $0 < a_{k+1} \le a_k \le 1/2$ and $1/2 \le 1 - a_k \le 1 - a_{k+1} < 1$ (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]

For every real $\eta > 0$ there is a natural $m \ge 1$ with $1/\iota(m) < \eta$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ).

[L3]

An affine map $t \mapsto ct + e$ of $\mathbb{R}$ is Lipschitz with constant $|c|$ , hence continuous, and a restriction to a metric subspace is continuous (Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction, Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent, Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ form, Absolute value in an ordered field, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L4]

A function whose restrictions to the members of a finite closed cover are continuous is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, claim 3, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, Continuity of a map of topological spaces at a point and globally).

[L5]

$I$ is a compact metric space, being closed in $\mathbb{R}$ and bounded (Heine-Borel in $\mathbb{R}^n$ : with the Euclidean metric a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, claim 3, Open cover, subcover, compact metric space, and compact subset of a metric space).

[L6]

Dini's theorem: on a compact metric space a sequence of continuous real functions with $g_k \le g_{k+1}$ pointwise, converging pointwise to a continuous $g$ , converges to $g$ uniformly (Dini's theorem: on a compact metric space a nondecreasing sequence of continuous real functions converging pointwise to a continuous limit converges uniformly, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum).

[L7]

Uniform convergence, its identification with convergence in the uniform metric, and the uniform limit theorem: a uniform limit of continuous functions is continuous (Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^{X}$ and on $C(X,Y)$ , Convergence in the uniform metric is exactly uniform convergence: one $N$ serving every point, For a nonempty set $X$ and a metric space $(Y,d)$ the uniform metric $\bar\rho(f,g) = \sup_{x} \min\{d(f(x),g(x)), 1\}$ is a metric on $Y^{X}$ , A uniform limit of continuous functions is continuous, so $C(X,Y)$ is closed in $Y^{X}$ under the uniform metric, claim 2).

[L8]

The ramps $r_k$ on $I$ are continuous, are pointwise nonincreasing, and converge pointwise to the indicator $\chi$ of $\{1\}$ , which is not continuous (Refuted: $C(X,Y)$ is closed in the topology of pointwise convergence. The ramps on $[0,1]$ converge pointwise to a discontinuous limit).

[L9]

The maximum and the minimum of a two-element set of reals exist and are among its elements (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

Verification

technique · direct

1.1

The two formulas for $s_k$ agree at $t = 1 - a_k$ , both giving $1 - a_k$ , and the closed sets $[0,1-a_k]$ and $[1-a_k,1]$ cover $I$ since $0 < 1 - a_k < 1$ ; each restriction is the restriction of an affine map of $\mathbb{R}$ , so $s_k$ is a well-defined continuous function on $I$ .

L1L3L4

1.2

$s_k(t) = \min\{t,\ 1-a_k\}$ for every $t \in I$ : for $t \le 1-a_k$ the minimum is $t$ , and for $t \ge 1-a_k$ it is $1-a_k$ .

L1L9

2.1

$s_k(t) \le s_{k+1}(t)$ for every $t$ and $k$ , since $1 - a_k \le 1 - a_{k+1}$ makes $\min\{t, 1-a_k\} \le \min\{t, 1-a_{k+1}\}$ ; with step 1.1 this is claim 1.

step 1.1step 1.2L1L9

2.2

$0 \le t - s_k(t) \le a_k$ for every $t \in I$ : for $t \le 1-a_k$ the difference is $0$ , and for $t > 1-a_k$ it is $t - (1-a_k) \le 1 - (1-a_k) = a_k$ .

step 1.2L1

3.1

Let $\varepsilon > 0$ be real; by [L2] there is a natural $m \ge 1$ with $1/\iota(m) < \varepsilon$ , and every $k \ge m$ has $a_k = 1/\iota(k+2) \le 1/\iota(m) < \varepsilon$ , so $|s_k(t) - t| \le a_k < \varepsilon$ for every $t \in I$ .

step 2.2L1L2

4.1

In particular $s_k(t) \to t$ for every $t \in I$ , and $\mathrm{id}_I$ is continuous, being the restriction of an affine map; this is claim 2.

step 3.1L3

5.1

$I$ is a compact metric space, every $s_k$ is continuous, the sequence is pointwise nondecreasing and its pointwise limit $\mathrm{id}_I$ is continuous, so Dini's theorem applies and $(s_k)$ converges to $\mathrm{id}_I$ uniformly; step 3.1 exhibits the same conclusion directly, an index $m$ serving every point at once.

step 2.1step 3.1step 4.1L5L6L7

6.1

For the failure clause, the ramps $r_k$ are continuous on the same compact $I$ and pointwise nonincreasing with pointwise limit the discontinuous $\chi$ ; were the convergence uniform, the limit would be continuous, so it is not uniform, and the conclusion of Dini's theorem fails for a family satisfying every one of its hypotheses except continuity of the limit.

step 5.1L5L7L8∎

Remarks

Dini's theorem is not needed for the positive half, and that is the point. Step 3.1 proves uniform convergence of $(s_k)$ by hand, because the discrepancy $t - s_k(t)$ is bounded by $a_k$ independently of $t$ . The example is worth stating because the general theorem gives the same conclusion from hypotheses that never mention a uniform bound: compactness, monotonicity, and continuity of the terms and of the limit.
Each hypothesis of Dini's theorem is doing something. Continuity of the limit fails for the ramps, and the conclusion fails with it. Compactness cannot be dropped either, though this page does not construct a witness for that. Monotonicity cannot be dropped: the moving spikes earlier on this page are continuous on the compact $I$ , converge pointwise to the continuous $\mathbf{0}$ , and do not converge uniformly, and they are not monotone at any point where the spike passes.
The nonincreasing form is the one the ramps illustrate, and it is the form obtained from Dini's theorem: on a compact metric space a nondecreasing sequence of continuous real functions converging pointwise to a continuous limit converges uniformly by applying it to the negatives of the functions, as that item's Statement records. Nothing here needs a separate proof.

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Sources

Dini's theorem (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 7 (standard reference, not scraped)

Dini's theorem applied to a nondecreasing sequence of piecewise linear approximations on [0,1][0,1][0,1], and what fails when the limit is not continuous