False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-05 (claude-sonnet-5)

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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.
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FALSE: a pointwise convergent sequence of continuous functions converges uniformly on every compact set

Statement

False claim: for metric spaces $X$ and $Y$ , if a sequence $(f_k)$ in $C(X,Y)$ converges pointwise to $f \in C(X,Y)$ (A sequence converges in the topology of pointwise convergence exactly when it converges at every point), then $(f_k)$ converges to $f$ uniformly on every compact subset of $X$ , that is $f_k \to f$ in the topology of compact convergence (The topology of compact convergence on $C(X,Y)$ for metric $X$ and $Y$ : uniform convergence on each compact subset of $X$ ).

The claim fails already on the compact space $X = [0,1]$ with $Y = \mathbb{R}$ , where it reduces to "pointwise convergence implies uniform convergence". The refutation below writes down the standard moving spike explicitly. The relation that is true is the inclusion of topologies (On $C(X,Y)$ with $X$ and $Y$ metric, uniform convergence is finer than compact convergence, which is finer than pointwise convergence): compact convergence implies pointwise convergence, and not the reverse.

No choice principle is used; every function below is given by a formula.

Facts & Assumptions

Given: The interval $X := [0,1] = \{\, t \in \mathbb{R} : 0 \le t \le 1 \,\}$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length) 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), the target $Y := \mathbb{R}$ with the same metric, the reals $a_k := 1/\iota(k+2)$ for $k \in \mathbb{N}$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field), and the constant function $\mathbf{0} : X \to \mathbb{R}$ with value $0$ .

[L1]

$\iota$ is strictly increasing on $\mathbb{N}$ and $\iota(n) > 0$ for $n \ge 1$ , so $0 < a_k \le 1/\iota(2) = 1/2$ and $0 < 2a_k \le 1$ for every $k \in \mathbb{N}$ , and $m \le n$ gives $a_n \le a_m$ (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 $\varepsilon > 0$ there is a natural $m \ge 1$ with $1/\iota(m) < \varepsilon$ (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 uniformly continuous, hence continuous; and the restriction of a continuous map 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, Isometry, isometric embedding, and the subspace metric on a subset, Absolute value in an ordered field).

[L4]

A function on a topological space 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).

[L5]

A subset of $\mathbb{R}$ is a compact subset exactly when it is 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, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L6]

A subset of a metric space is closed exactly when its complement is open, and a set is open exactly when each of its points has a ball around it inside the set (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

[L7]

The basic sets of the topology of compact convergence are $B_K(f,\varepsilon) = \{\, g : d(f(x),g(x)) < \varepsilon \text{ for every } x \in K \,\}$ , and a sequence converging to $f$ in a topology is eventually inside every neighbourhood of $f$ (The topology of compact convergence on $C(X,Y)$ for metric $X$ and $Y$ : uniform convergence on each compact subset of $X$ , The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C(X,Y)$ ).

Refutation

technique · direct

1.1

$[0,1]$ is bounded, being contained in the ball $B(0,2)$ of $\mathbb{R}$ , and closed in $\mathbb{R}$ , since a point $y < 0$ has $B(y,-y)$ inside the complement and a point $y > 1$ has $B(y, y-1)$ inside the complement; so $[0,1]$ is a compact subset of $\mathbb{R}$ and $(X,d)$ is a compact metric space.

L5L6

1.2

For $k \in \mathbb{N}$ define $f_k : X \to \mathbb{R}$ by $f_k(t) := t/a_k$ for $0 \le t \le a_k$ , by $f_k(t) := 2 - t/a_k$ for $a_k \le t \le 2a_k$ , and by $f_k(t) := 0$ for $2a_k \le t \le 1$ .

constructL1

2.1

The three formulas agree where their domains overlap: at $t = a_k$ both of the first two give $1$ , and at $t = 2a_k$ both of the last two give $0$ ; so $f_k$ is a well-defined function on $X$ , the three closed sets $[0,a_k]$ , $[a_k,2a_k]$ and $[2a_k,1]$ covering $X$ because $0 < 2a_k \le 1$ .

step 1.2L1

2.2

$f_k(0) = 0$ for every $k$ , from the first formula.

step 1.2

2.3

For $t \in X$ with $t > 0$ : by [L2] there is a natural $m \ge 1$ with $1/\iota(m) < t/2$ , and then every $k \ge m$ has $k + 2 > m$ , hence $a_k = 1/\iota(k+2) \le 1/\iota(m) < t/2$ , hence $2a_k < t$ and $f_k(t) = 0$ by the third formula.

step 1.2L1L2

2.4

On the other hand $f_k(a_k) = a_k/a_k = 1$ for every $k \in \mathbb{N}$ , and $a_k \in X$ because $0 < a_k \le 1/2 \le 1$ .

step 1.2L1

3.1

Each of the three restrictions is the restriction of an affine map of $\mathbb{R}$ , hence continuous; so $f_k$ is continuous on $X$ by the pasting lemma for a finite closed cover, and $f_k \in C(X,\mathbb{R})$ .

step 1.2step 2.1L3L4

3.2

By steps 2.2 and 2.3 the sequence $(f_k(t))$ is eventually $0$ for every $t \in X$ , so $f_k(t) \to 0 = \mathbf{0}(t)$ for every $t \in X$ ; that is, $(f_k)$ converges pointwise to $\mathbf{0}$ , which is continuous, being constant.

step 2.2step 2.3L3

3.3

Hence for every $k \in \mathbb{N}$ the value $|f_k(a_k) - \mathbf{0}(a_k)| = 1$ is not below $1/2$ , so $f_k \notin B_{X}(\mathbf{0}, 1/2)$ , while $B_{X}(\mathbf{0},1/2)$ is a basic open set of the topology of compact convergence containing $\mathbf{0}$ , the whole space $X$ being compact by step 1.1.

step 1.1step 2.4L7

4.1

So no tail of $(f_k)$ lies in the neighbourhood $B_X(\mathbf{0},1/2)$ of $\mathbf{0}$ : the sequence does not converge to $\mathbf{0}$ in the topology of compact convergence, although by step 3.2 it converges to $\mathbf{0}$ pointwise.

step 3.2step 3.3L7

5.1

The pair $(X,Y) = ([0,1],\mathbb{R})$ with the sequence $(f_k)$ and the limit $\mathbf{0}$ therefore satisfies the hypothesis of the claim and violates its conclusion at the compact set $K = X$ , so the claim is false.

step 3.2step 4.1∎

Remarks

The failure is not about the size of the domain. The domain here is compact, so "uniformly on every compact set" is the same as "uniformly", and the witness shows that pointwise convergence does not give uniform convergence even there. What moves is the place where the two functions differ: the spike has height $1$ for every $k$ and merely slides towards $0$ .
The area under the spike does tend to $0$ , so this witness does not also separate the integral from its pointwise limit: the standard warning that pointwise convergence controls no integral needs a spike whose height grows as its base shrinks. Nothing about integration is claimed here.
What is true in this direction. Uniform convergence implies convergence on every compact set, which implies pointwise convergence (On $C(X,Y)$ with $X$ and $Y$ metric, uniform convergence is finer than compact convergence, which is finer than pointwise convergence); the reverse of each implication fails, and the companion page separates the two rightmost topologies with a different witness on $\mathbb{R}$ .
The index shift is not cosmetic. $\mathbb{N}$ contains $0$ , so the spike is built on $1/\iota(k+2)$ and not on $1/k$ : at $k = 0$ the reciprocal $1/\iota(1)$ would give a support $[0,2]$ reaching outside $[0,1]$ , and the pasting lemma would have nothing to paste.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 173 results over 27 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

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