CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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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.
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Refuted: convergence uniformly on every compact subset of $\mathbb{R}$ implies uniform convergence. The maps $x \mapsto x/(n+1)$ separate the two

Statement refuted

Refuted claim: if a sequence $(g_k)$ in $C(X,\mathbb{R})$ converges to $g$ uniformly on every compact subset of $X$ , that is 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 $X$ ), then it converges to $g$ uniformly (Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^{X}$ and on $C(X,Y)$ ).

The witness is $X = \mathbb{R}$ with its usual metric and

$g_k(x) := \frac{x}{\iota(k+1)} \qquad (k \in \mathbb{N},\ x \in \mathbb{R}),$

$\iota$ being the canonical natural of $\mathbb{R}$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field), so that $\iota(k+1) \ge 1 > 0$ and the quotient is defined. These converge to the constant function $\mathbf{0}$ uniformly on every compact subset of $\mathbb{R}$ and satisfy $\bar\rho(g_k,\mathbf{0}) = 1$ for every $k$ , so they do not converge uniformly.

This is the strictness of the right-hand inclusion of On $C(X,Y)$ with $X$ and $Y$ metric, uniform convergence is finer than compact convergence, which is finer than pointwise convergence; the left-hand one is separated on the companion example of this page. Note that the domain has to be non-compact for such a witness to exist, since on a compact domain the two topologies coincide.

Facts & Assumptions

Given: $\mathbb{R}$ with the usual metric $d(s,t) = |s-t|$ (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), the maps $g_k(x) = x/\iota(k+1)$ , the constant function $\mathbf{0}$ , and the truncated metric $\bar d = \min\{d,1\}$ with the uniform metric $\bar\rho$ on $\mathbb{R}^{\mathbb{R}}$ (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 $Y^{X}$ ).

[L1]

$\iota$ is strictly increasing on $\mathbb{N}$ with $\iota(n) > 0$ for $n \ge 1$ , and $0 < u \le v$ gives $0 < 1/v \le 1/u$ (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]

A map $x \mapsto cx$ of $\mathbb{R}$ is Lipschitz with constant $|c|$ , hence continuous, and $|cx| = |c||x|$ (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, Continuity of a map of topological spaces at a point and globally, 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).

[L4]

A compact subset of $\mathbb{R}$ is closed and bounded, and a bounded subset lies in some ball $B(0,r)$ , so $|x| < r$ for all its points (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, Open ball, closed ball and sphere in a metric space, 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, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L5]

The sets $B_K(f,\varepsilon)$ centred at $f$ form a neighbourhood base at $f$ in the topology of compact convergence, and $B_K(f,\varepsilon) = \{\, h : |f(x)-h(x)| < \varepsilon$ for every $x \in K \,\}$ (The topology of compact convergence on $C(X,Y)$ for metric $X$ and $Y$ : uniform convergence on each compact subset of $X$ , fact (U4)).

[L6]

$\bar d \le d$ ; $\bar d(u,v) = |u-v|$ whenever $|u-v| \le 1$ ; $\bar\rho(f,h)$ is the least upper bound of $\{\, \bar d(f(x),h(x)) : x \in \mathbb{R} \,\}$ ; and convergence in $\bar\rho$ is uniform convergence ( $\min(d,1)$ and $d/(1+d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology, 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 $Y^{X}$ , Convergence in the uniform metric is exactly uniform convergence: one $N$ serving every point, Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set, Complete ordered field (least-upper-bound property), Suprema and infima are unique).

Counterexample

technique · direct

1.1

Each $g_k$ is continuous, being Lipschitz with constant $1/\iota(k+1)$ , so $g_k \in C(\mathbb{R},\mathbb{R})$ ; and $\mathbf{0}$ is continuous, being constant.

L1L3

1.2

Let $K \subseteq \mathbb{R}$ be compact and let $\varepsilon > 0$ be real; $K$ is bounded, so fix a real $r > 0$ with $|x| < r$ for every $x \in K$ .

L4choose

1.3

On the other hand, for each $k \in \mathbb{N}$ the point $x_k := \iota(k+1)$ satisfies $|g_k(x_k) - \mathbf{0}(x_k)| = \iota(k+1)/\iota(k+1) = 1$ , so $\bar d(g_k(x_k),\mathbf{0}(x_k)) = 1$ and therefore $\bar\rho(g_k,\mathbf{0}) \ge 1$ ; since $\bar\rho \le 1$ always, $\bar\rho(g_k,\mathbf{0}) = 1$ .

L1L6

2.1

By [L2] fix a natural $m \ge 1$ with $1/\iota(m) < \varepsilon/r$ .

step 1.2L2choose

2.2

So no index $K_0$ makes $\bar\rho(g_k,\mathbf{0}) < 1/2$ for all $k \ge K_0$ , and $(g_k)$ does not converge to $\mathbf{0}$ in the uniform metric, that is not uniformly.

step 1.3L6

3.1

For every $k \ge m$ and every $x \in K$ : $|g_k(x) - \mathbf{0}(x)| = |x|/\iota(k+1) < r/\iota(k+1) \le r/\iota(m) < \varepsilon$ , using $k+1 > m$ and the monotonicity of $\iota$ and of reciprocals.

step 1.2step 2.1L1L3

4.1

Hence $g_k \in B_K(\mathbf{0},\varepsilon)$ for every $k \ge m$ ; as $K$ and $\varepsilon$ were arbitrary and the sets $B_K(\mathbf{0},\varepsilon)$ are a neighbourhood base at $\mathbf{0}$ , the sequence $(g_k)$ converges to $\mathbf{0}$ in the topology of compact convergence.

step 3.1L5

5.1

The sequence $(g_k)$ therefore satisfies the hypothesis of the claim and violates its conclusion, so the claim is false.

step 4.1step 2.2∎

Remarks

What goes wrong is arbitrarily far out. On any fixed bounded region the maps $g_k$ do flatten to $0$ , and a compact subset of $\mathbb{R}$ is bounded; the discrepancy $|g_k(x)| = |x|/(k+1)$ reaches $1$ only at $x = k+1$ , which escapes every compact set as $k$ grows. Uniform convergence asks for control at every point at once, including those.
The same family shows the two topologies are different as topologies, not merely that one sequence behaves differently in them: the difference is already visible in a basic neighbourhood, since $B_{\bar\rho}(\mathbf{0},1/2)$ contains no $g_k$ while every compact-convergence neighbourhood of $\mathbf{0}$ contains a tail of them.
The index shift is the usual one. $\mathbb{N}$ contains $0$ , so the family is written with $\iota(k+1)$ ; at $k = 0$ this is $1$ and $g_0$ is the identity, which is exactly the intended first term. Writing $x/\iota(k)$ would divide by $0$ at $k = 0$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 167 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

Compact convergence (Wikipedia) (standard reference, not scraped)
Uniform convergence (Wikipedia) (standard reference, not scraped)

Refuted: convergence uniformly on every compact subset of R\mathbb{R}R implies uniform convergence. The maps x↦x/(n+1)x \mapsto x/(n+1)x↦x/(n+1) separate the two