CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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

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 $Z = R$ with its usual topology, $Y_{0} = {a, b}$ with $a \neq b$ 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

$E (f, g) = Q_{R},$

which is dense in $R$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets) and is not all of $R$ , 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: $R$ with its usual topology; the set $Q_{R}$ of rationals inside $R$ ; the two-point set $Y_{0} = {a, b}$ with $a \neq b$ and the indiscrete topology; and the maps $f \equiv a$ and $g$ equal to $a$ on $Q_{R}$ and to $b$ off it.

[A1]

$U \subseteq R$ 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 $R$ 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 $R$ : the nine order-convex forms, nondegeneracy, and length).

[A2]

The indiscrete topology on $Y_{0}$ is ${\emptyset, Y_{0}}$ , 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]

$A \subseteq R$ is dense exactly when $U \cap A \neq \emptyset$ for every nonempty open $U$ , equivalently when $\overline{A} = R$ (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 $f, g : R \to Y_{0}$ with $f (x) = a$ for every $x$ , $g (x) = a$ for $x \in Q_{R}$ and $g (x) = b$ otherwise.

A2construct

1.2

$Q_{R}$ is dense in $R$ : 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 $t \notin Q_{R}$ .

L3choose

2.1

Both maps are continuous: the only open subsets of $Y_{0}$ are $\emptyset$ and $Y_{0}$ , whose preimages are $\emptyset$ and $R$ , both open.

step 1.1A2L1

2.2

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

step 1.1

3.1

$\overline{E (f, g)} = R$ by steps 1.2 and 2.2, while $E (f, g) \neq R$ since $t \notin Q_{R}$ 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 $R$ can be an agreement set here. Replacing $Q_{R}$ by an arbitrary $S \subseteq R$ 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

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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}R→{a,b} into the indiscrete two-point space have agreement set QQ