CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)

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Refuted: $C (X, Y)$ is closed in the topology of pointwise convergence. The ramps on $[0, 1]$ converge pointwise to a discontinuous limit

Statement refuted

Refuted claim: for a topological space $X$ and a metric space $Y$ the set $C (X, Y)$ is closed in $Y^{X}$ for the topology of pointwise convergence (The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ ). In particular, this would imply that every pointwise limit of a sequence of continuous functions is continuous.

The witness is the sequence of ramps on $I : = [0, 1]$ . With $a_{k} : = 1 / ι (k + 2)$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field), so that $0 < a_{k} \leq 1 / 2$ , define $r_{k} : I \to R$ by

$r_{k} (t) : = 0 (0 \leq t \leq 1 - a_{k}), r_{k} (t) : = \frac{t - (1 - a_{k})}{a_{k}} (1 - a_{k} \leq t \leq 1) .$

Each $r_{k}$ is continuous, the sequence $(r_{k})$ converges pointwise to the indicator function

$χ (t) : = 0 (t < 1), χ (1) : = 1,$

and $χ$ is not continuous at $1$ . So $C (I, R)$ is not closed in $R^{I}$ for the topology of pointwise convergence.

The sequence is moreover pointwise nonincreasing, $r_{k + 1} (t) \leq r_{k} (t)$ for every $t \in I$ and every $k \in N$ (step 2.1 below). That is recorded here because it is the configuration Dini's theorem rules out on a compact domain when the limit is continuous; here the limit is not continuous, and the conclusion of Dini's theorem fails.

This is exactly what the uniform topology repairs. For the uniform metric $C (X, Y)$ is closed (A uniform limit of continuous functions is continuous, so $C (X, Y)$ is closed in $Y^{X}$ under the uniform metric, claim 3), so the convergence above cannot be uniform. Explicitly, for $0 < η < 1$ , take $t = 1 - η a_{k} < 1$ . Then $χ (t) = 0$ and $r_{k} (t) = 1 - η$ . Since both functions take values in $[0, 1]$ , this proves $\sup_{t \in I} ∣ r_{k} (t) - χ (t) ∣ = 1$ for every $k$ . The supremum is not attained: below $1$ the ramp is less than $1$ , and at $1$ the difference is zero.

Facts & Assumptions

Given: $I = [0, 1]$ with the metric $d (s, t) = ∣ s - t ∣$ inherited from $R$ (The absolute value makes $R$ 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 $R$ : the nine order-convex forms, nondegeneracy, and length), the target $R$ with the same metric, the reals $a_{k} = 1 / ι (k + 2)$ , the ramps $r_{k}$ and the indicator $χ$ displayed above.

[L1]

$ι$ is strictly increasing on $N$ with $ι (n) > 0$ for $n \geq 1$ , and $0 < u \leq v$ gives $0 < 1 / v \leq 1 / u$ ; hence $0 < a_{k} \leq 1 / ι (2) = 1 / 2$ and $1 / 2 \leq 1 - a_{k} < 1$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

For every real $η > 0$ there is a natural $m \geq 1$ with $1 / ι (m) < η$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L3]

An affine map $t \mapsto c t + e$ of $R$ is Lipschitz with constant $∣ c ∣$ , hence continuous, and the restriction of a continuous map to a metric subspace is continuous (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , 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 $ε$ - $δ$ form, Absolute value in an ordered field, Isometry, isometric embedding, and the subspace metric on a subset, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

[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]

A sequence converges in the topology of pointwise convergence exactly when it converges at every point (A sequence converges in the topology of pointwise convergence exactly when it converges at every point, Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure).

[L6]

$h : I \to R$ is continuous at $a$ exactly when for every real $ε > 0$ there is a real $δ > 0$ with $∣ h (t) - h (a) ∣ < ε$ whenever $t \in I$ and $∣ t - a ∣ < δ$ (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, 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, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L7]

A two-element set of reals has a maximum and a minimum, each of which is one of the two elements (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

Counterexample

technique · direct

1.1

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

L1L3L4

1.2

$r_{k} (1) = (1 - (1 - a_{k})) / a_{k} = 1$ for every $k$ .

1.3

Let $t \in I$ with $t < 1$ ; by [L2] there is a natural $m \geq 1$ with $1 / ι (m) < 1 - t$ , and then every $k \geq m$ has $a_{k} = 1 / ι (k + 2) \leq 1 / ι (m) < 1 - t$ , hence $t < 1 - a_{k}$ and $r_{k} (t) = 0$ .

L1L2

1.4

$χ$ is not continuous at $1$ : take $ε : = 1 / 2$ and let $δ > 0$ be any real. Put $s : = 1 - \min {δ / 2, 1 / 2}$ . Then $1 / 2 \leq s < 1$ and $∣ s - 1 ∣ = \min {δ / 2, 1 / 2} \leq δ / 2 < δ$ , yet $∣ χ (s) - χ (1) ∣ = 1$ , which is not below $1 / 2$ .

L6L7

2.1

$(r_{k})$ is pointwise nonincreasing: for $t \leq 1 - a_{k + 1}$ one has $r_{k + 1} (t) = 0 \leq r_{k} (t)$ , the values of $r_{k}$ being nonnegative; and for $t > 1 - a_{k + 1}$ , which forces $t > 1 - a_{k}$ since $a_{k + 1} < a_{k}$ , writing $u : = 1 - t$ with $0 \leq u < a_{k + 1}$ gives $r_{k} (t) = 1 - u / a_{k}$ and $r_{k + 1} (t) = 1 - u / a_{k + 1}$ , and $a_{k + 1} < a_{k}$ gives $u / a_{k} \leq u / a_{k + 1}$ , hence $r_{k + 1} (t) \leq r_{k} (t)$ . This includes $t = 1$ , where $u = 0$ and both ramps equal $1$ .

step 1.1L1

2.2

By steps 1.2 and 1.3 the sequence $(r_{k} (t))$ is eventually equal to $χ (t)$ for every $t \in I$ , so $r_{k} (t) \to χ (t)$ for every $t$ , and therefore $r_{k} \to χ$ in the topology of pointwise convergence on $R^{I}$ .

step 1.2step 1.3L5

3.1

So $χ \notin C (I, R)$ although $χ$ is a limit in the topology of pointwise convergence of a sequence in $C (I, R)$ ; hence $C (I, R)$ is not closed in that topology, and the claim is false.

step 1.1step 2.2step 1.4∎

Remarks

Monotonicity is not what fails. Step 2.1 shows the ramps decrease pointwise to $χ$ on the compact domain $I$ , with every $r_{k}$ continuous; the only hypothesis of Dini's theorem that is missing is continuity of the limit, and its conclusion, uniform convergence, fails. The last example on this page uses this family for exactly that contrast.
Closedness in the pointwise topology is not a mild question. The set $C (I, R)$ is in fact dense in $R^{I}$ for the topology of pointwise convergence, since a basic neighbourhood constrains only finitely many values and any finite list of values is realised by a continuous function. Nothing above needs that, and it is not proved here.
What survives is the uniform statement. Convergence in the uniform metric does force continuity of the limit (A uniform limit of continuous functions is continuous, so $C (X, Y)$ is closed in $Y^{X}$ under the uniform metric), so the failure above is a failure of the topology, not of the limit operation: the same sequence, tested against a stronger notion of convergence, simply does not converge.

Depends on

Used by

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

Dependency tree · two levels

78 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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

Refuted: C(X,Y)C(X,Y) is closed in the topology of pointwise convergence. The ramps on [0,1][0,1] converge pointwise to a discontinuous limit