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

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Refuted: a pointwise bounded family of continuous functions is equicontinuous. The spikes are bounded by $1$ everywhere and are not equicontinuous at $0$

Statement refuted

Refuted claim: a pointwise bounded family of continuous maps between metric spaces is equicontinuous (Equicontinuity at a point, uniform equicontinuity, and pointwise boundedness of a family of maps between metric spaces).

The witness is the family of moving spikes on $I := [0,1]$ already built in FALSE: a pointwise convergent sequence of continuous functions converges uniformly on every compact set: with $a_k := 1/\iota(k+2)$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field),

$f_k(t) = \frac{t}{a_k} \ \ (0 \le t \le a_k), \qquad f_k(t) = 2 - \frac{t}{a_k} \ \ (a_k \le t \le 2a_k), \qquad f_k(t) = 0 \ \ (2a_k \le t \le 1) .$

Every $f_k$ is continuous and takes values in $[0,1]$ , so $\mathcal{F} := \{\, f_k : k \in \mathbb{N} \,\}$ is pointwise bounded; but $\mathcal{F}$ is not equicontinuous at $0$ , because $f_k$ climbs from $0$ to $1$ over an interval of length $a_k$ , and $a_k$ can be made smaller than any prescribed $\delta$ .

Facts & Assumptions

Given: $I = [0,1]$ with the metric $d(s,t) = |s-t|$ inherited from $\mathbb{R}$ (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 target $\mathbb{R}$ with the same metric, the reals $a_k = 1/\iota(k+2)$ , the spikes $f_k$ displayed above and the family $\mathcal{F} = \{\, f_k : k \in \mathbb{N} \,\}$ .

[L1]

Each $f_k$ is a well-defined continuous function $I \to \mathbb{R}$ , $f_k(0) = 0$ , $f_k(a_k) = 1$ , and $0 < a_k \le 1/2$ , so $a_k \in I$ (FALSE: a pointwise convergent sequence of continuous functions converges uniformly on every compact set, Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, The canonical natural $\iota(n) = n \cdot 1_F$ of a 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).

[L2]

$0 \le f_k(t) \le 1$ for every $t \in I$ and every $k$ : the three formulas take values $t/a_k \in [0,1]$ , $2 - t/a_k \in [0,1]$ and $0$ respectively on their pieces (Maximum and minimum of a set, Every nonempty finite set of reals has a maximum and a minimum, Absolute value in an ordered field).

[L3]

A family $\mathcal{F}$ is pointwise bounded when for each $t$ the set $\{\, f(t) : f \in \mathcal{F} \,\}$ lies in some ball of the target, and equicontinuous at $a$ when for every real $\varepsilon > 0$ there is a real $\delta > 0$ with $|f(t)-f(a)| < \varepsilon$ for every $f \in \mathcal{F}$ and every $t$ with $|t-a| < \delta$ (Equicontinuity at a point, uniform equicontinuity, and pointwise boundedness of a family of maps between metric spaces, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open ball, closed ball and sphere in a metric space, Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ form).

[L4]

For every real $\eta > 0$ there is a natural $m \ge 1$ with $1/\iota(m) < \eta$ ; $\iota$ is strictly increasing with $\iota(n) > 0$ for $n \ge 1$ ; and $0 < u \le v$ gives $0 < 1/v \le 1/u$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

Counterexample

technique · direct

1.1

For every $t \in I$ the set $\{\, f_k(t) : k \in \mathbb{N} \,\}$ is contained in $[0,1]$ and hence in the ball $B(0,2)$ of $\mathbb{R}$ , so $\mathcal{F}$ is pointwise bounded.

L2L3

1.2

Every member of $\mathcal{F}$ is continuous.

1.3

Take $\varepsilon := 1/2$ and let $\delta > 0$ be any real; by [L4] there is a natural $m \ge 1$ with $1/\iota(m) < \delta$ , and setting $k := m$ gives $a_k = 1/\iota(m+2) \le 1/\iota(m) < \delta$ , since $m + 2 > m$ and $\iota$ is increasing.

L4choose

2.1

For that $k$ the point $a_k$ lies in $I$ and satisfies $|a_k - 0| = a_k < \delta$ , while $|f_k(a_k) - f_k(0)| = |1 - 0| = 1$ , which is not below $\varepsilon = 1/2$ .

step 1.3L1

3.1

So no $\delta > 0$ serves the whole family at $\varepsilon = 1/2$ and the point $0$ : the family $\mathcal{F}$ is not equicontinuous at $0$ , hence not equicontinuous.

step 1.3step 2.1L3

4.1

By steps 1.1, 1.2 and 3.1 the family $\mathcal{F}$ is a pointwise bounded family of continuous functions that is not equicontinuous, so the claim is false.

step 1.1step 1.2step 3.1∎

Remarks

The values stay in $[0,1]$ and the slopes do not. $f_k$ is Lipschitz with constant $1/a_k = \iota(k+2)$ and with no smaller one, so the family has no common Lipschitz constant. That is the contrast with the previous example on this page, where fixing the constant at $1$ is exactly what produced uniform equicontinuity.
The failure is at one point only, and that is enough. The family is equicontinuous at every $t > 0$ : for $0 < \delta < t/2$ and $k$ large the spike is identically $0$ on the interval around $t$ , and the finitely many remaining members are individually continuous. Equicontinuity is required at every point, so failure at $0$ refutes the claim.
Both hypotheses of an Ascoli-type theorem are therefore needed, and neither implies the other: this family is pointwise bounded and not equicontinuous, and the $1$ -Lipschitz maps of the previous example are equicontinuous and not pointwise bounded.

Depends on

Used by

Nothing in the library uses this result yet.

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Direct dependencies and their dependencies through the next three levels: 131 results over 23 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

Refuted: a pointwise bounded family of continuous functions is equicontinuous. The spikes are bounded by 111 everywhere and are not equicontinuous at 000