CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

How statement and proof provenance work

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: the agreement set of two continuous maps is closed, with no hypothesis on the codomain. Two continuous maps $\mathbb{R} \to \{a,b\}$ into the indiscrete two-point space have agreement set $\mathbb{Q}$

Statement refuted

False claim: for continuous maps $f, g : Z \to Y$ between topological spaces the agreement set $E(f,g) = \{\, z \in Z : f(z) = g(z) \,\}$ is closed in $Z$ , with no hypothesis on the codomain $Y$ .

The witness is the pair of maps of FALSE: two continuous maps that agree on a dense subset of their common domain are equal, with no hypothesis on the codomain: take $Z = \mathbb{R}$ with its usual topology, $Y_0 = \{a,b\}$ with $a \ne b$ and the indiscrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), $f \equiv a$ , and $g$ equal to $a$ at every rational and to $b$ at every irrational. Both are continuous, and

$E(f,g) \;=\; \mathbb{Q}_{\mathbb{R}} ,$

which is dense in $\mathbb{R}$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets) and is not all of $\mathbb{R}$ , hence is not closed.

So the hypothesis dropped is the Hausdorff condition on the codomain, which is what For continuous $f, g : Z \to Y$ with $Y$ Hausdorff the agreement set $\{ z \in Z : f(z) = g(z) \}$ is closed in $Z$ assumes; and the failure is the worst possible one, the agreement set being dense rather than merely non-closed.

Facts & Assumptions

Given: $\mathbb{R}$ with its usual topology; the set $\mathbb{Q}_{\mathbb{R}}$ of rationals inside $\mathbb{R}$ ; the two-point set $Y_0 = \{a,b\}$ with $a \ne b$ and the indiscrete topology; and the maps $f \equiv a$ and $g$ equal to $a$ on $\mathbb{Q}_{\mathbb{R}}$ and to $b$ off it.

[A1]

$U \subseteq \mathbb{R}$ is open in the usual topology exactly when for every $x \in U$ there is a real $r > 0$ with $(x - r, x + r) \subseteq U$ (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, claims 2 and 3, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[A2]

The indiscrete topology on $Y_0$ is $\{\varnothing, Y_0\}$ , and the two-point indiscrete space is not Hausdorff (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[A3]

$A \subseteq \mathbb{R}$ is dense exactly when $U \cap A \ne \varnothing$ for every nonempty open $U$ , equivalently when $\overline{A} = \mathbb{R}$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets, forms 1 and 2).

[L1]

A function between topological spaces is continuous exactly when the preimage of every open set is open (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\overline{A}) \subseteq \overline{f(A)}$ , clauses (a) and (b), Continuity of a map of topological spaces at a point and globally).

[L2]

Strictly between any two reals lies a rational (The rationals embed densely in the reals).

[L3]

The set of irrationals is uncountable, hence not finite, hence nonempty (The irrationals are uncountable, Finite, countably infinite, countable, uncountable).

[L4]

A set is closed exactly when it equals its own closure (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set, claim 2).

[L5]

If the codomain is Hausdorff then the agreement set of two continuous maps is closed, and two such maps agreeing on a dense subset are equal (For continuous $f, g : Z \to Y$ with $Y$ Hausdorff the agreement set $\{ z \in Z : f(z) = g(z) \}$ is closed in $Z$ , Two continuous maps into a Hausdorff space that agree on a dense subset are equal).

Counterexample

technique · constructive

1.1

Give $Y_0 = \{a,b\}$ the indiscrete topology and take $f, g : \mathbb{R} \to Y_0$ with $f(x) = a$ for every $x$ , $g(x) = a$ for $x \in \mathbb{Q}_{\mathbb{R}}$ and $g(x) = b$ otherwise.

A2construct

1.2

$\mathbb{Q}_{\mathbb{R}}$ is dense in $\mathbb{R}$ : given a nonempty open $U$ , pick $x \in U$ and by [A1] a real $r > 0$ with $(x - r, x + r) \subseteq U$ ; by [L2] some rational lies strictly between $x - r$ and $x + r$ , hence in $U$ .

A1A3L2

1.3

There is a real $t \notin \mathbb{Q}_{\mathbb{R}}$ .

L3choose

2.1

Both maps are continuous: the only open subsets of $Y_0$ are $\varnothing$ and $Y_0$ , whose preimages are $\varnothing$ and $\mathbb{R}$ , both open.

step 1.1A2L1

2.2

$E(f,g) = \mathbb{Q}_{\mathbb{R}}$ : for $x \in \mathbb{Q}_{\mathbb{R}}$ both maps take the value $a$ , and for $x \notin \mathbb{Q}_{\mathbb{R}}$ they take the values $a$ and $b$ , which differ.

step 1.1

3.1

$\overline{E(f,g)} = \mathbb{R}$ by steps 1.2 and 2.2, while $E(f,g) \ne \mathbb{R}$ since $t \notin \mathbb{Q}_{\mathbb{R}}$ by step 1.3; so $E(f,g)$ is not closed by [L4].

step 1.2step 1.3step 2.2A3L4

4.1

Steps 2.1 and 3.1 exhibit two continuous maps whose agreement set is not closed, so the claim is false; by [A2] the codomain is not Hausdorff, which is exactly the hypothesis [L5] carries.

step 2.1step 3.1A2L5discharge-construct∎

Remarks

This is the sharp form of the failure recorded by FALSE: two continuous maps that agree on a dense subset of their common domain are equal, with no hypothesis on the codomain. That item refutes the equality of two maps agreeing on a dense set; the present one locates the reason, namely that without a separation hypothesis on the codomain the agreement set need not be closed, so a dense agreement set need not be all of the domain. Both failures come from the same pair of maps.
Any subset of $\mathbb{R}$ can be an agreement set here. Replacing $\mathbb{Q}_{\mathbb{R}}$ by an arbitrary $S \subseteq \mathbb{R}$ in the definition of $g$ gives $E(f,g) = S$ by the computation of step 2.2, and $g$ is still continuous by step 2.1. So into a two-point indiscrete codomain the agreement set carries no topological information whatsoever, which is the strongest way the claim can fail.
Why a two-point indiscrete codomain is the natural witness. A space with at most one point is Hausdorff vacuously (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not), so a witness needs at least two points; and the indiscrete topology is the coarsest topology on a two-point set, which is what makes every map into it continuous and so removes any need to verify continuity. Other non-Hausdorff topologies on two points exist — Sierpinski space is one (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies) — and would require that verification instead.

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: 147 results over 36 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

Trivial topology (Wikipedia) (standard reference, not scraped)
Dense set (Wikipedia) (standard reference, not scraped)
Hausdorff space (Wikipedia) (standard reference, not scraped)

Refuted: the agreement set of two continuous maps is closed, with no hypothesis on the codomain. Two continuous maps R→{a,b}\mathbb{R} \to \{a,b\}R→{a,b} into the indiscrete two-point space have agreement set Q\mathbb{Q}Q