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

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FALSE: two continuous maps that agree on a dense subset of their common domain are equal, with no hypothesis on the codomain

Statement

False claim: if $Z$ and $Y$ are topological spaces, $D \subseteq Z$ is dense (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets), and $f, g : Z \to Y$ are continuous (Continuity of a map of topological spaces at a point and globally) with $f (d) = g (d)$ for every $d \in D$ , then $f = g$ .

The refutation takes $Z = R$ with its usual topology (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, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), the dense set $D = Q_{R}$ of rationals inside $R$ (The rationals embed densely in the reals), and for codomain the two-point set $Y_{0} = {a, b}$ with $a \neq b$ carrying the indiscrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), which is not Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not). The maps are the constant $f \equiv a$ and the map $g$ equal to $a$ at every rational and to $b$ at every irrational. Both are continuous, they agree on $D$ , and they differ at every irrational.

The hypothesis the claim drops is the Hausdorff condition on the codomain, which is what Two continuous maps into a Hausdorff space that agree on a dense subset are equal assumes.

Facts & Assumptions

Given: $R$ with its usual topology; the set $Q_{R}$ of rationals inside $R$ ; and a two-element set $Y_{0} = {a, b}$ with $a \neq b$ .

[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 ${\emptyset, Y_{0}}$ ; the two-point indiscrete space is not Hausdorff, the only open set containing either point being $Y_{0}$ (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, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[A3]

$A \subseteq R$ is dense exactly when $U \cap A \neq \emptyset$ for every nonempty open $U \subseteq R$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets, form 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 $R ∖ Q_{R}$ of irrationals is uncountable (The irrationals are uncountable); an uncountable set is not finite and the empty set is finite, so it is nonempty (Finite, countably infinite, countable, uncountable).

[L4]

Two continuous maps into a Hausdorff space agreeing on a dense subset of their common domain are equal (Two continuous maps into a Hausdorff space that agree on a dense subset are equal).

Refutation

technique · constructive

1.1

Give $Y_{0} = {a, b}$ the indiscrete topology; it is not Hausdorff.

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 $t \notin Q_{R}$ , the set of irrationals being nonempty.

L3choose

2.1

Every function $h$ from a topological space into $Y_{0}$ is continuous: the only open subsets of $Y_{0}$ are $\emptyset$ and $Y_{0}$ , whose preimages are $\emptyset$ and the whole domain, both open.

step 1.1A2L1

3.1

Define $f : R \to Y_{0}$ by $f (x) : = a$ for every $x$ , and $g : R \to Y_{0}$ by $g (x) : = a$ for $x \in Q_{R}$ and $g (x) : = b$ otherwise; both are continuous by step 2.1.

step 2.1construct

4.1

$f$ and $g$ agree at every point of $Q_{R}$ , which is dense in $R$ by step 1.2.

step 1.2step 3.1

4.2

$f (t) = a$ and $g (t) = b$ for the irrational $t$ of step 1.3, and $a \neq b$ , so $f \neq g$ .

step 1.3step 3.1

5.1

Steps 3.1, 4.1 and 4.2 exhibit two continuous maps agreeing on a dense subset of their common domain and not equal, so the claim is false; by step 1.1 the codomain is not Hausdorff, which is exactly the hypothesis [L4] carries and the claim drops.

step 1.1step 3.1step 4.1step 4.2L4discharge-construct∎

Remarks

The failure is as large as it can be. The two maps agree precisely on $Q_{R}$ and differ at every other point of $R$ , so nothing is salvaged by weakening the conclusion from equality to agreement off a small set: the disagreement set is the whole of the irrationals.
Continuity is not being cheated. Both maps are continuous for the honest reason recorded in step 2.1, that the codomain has only two open sets. No pathology of $R$ is involved, and the same construction runs with $R$ replaced by any space with a dense subset that is not the whole space.
Why a two-point codomain suffices. The Hausdorff condition is a statement about pairs of distinct points, so the smallest space that can fail it has two points, and the indiscrete topology is the coarsest topology on it. Taking the coarsest topology is also what makes every map into it continuous, so the witness needs no verification of continuity beyond counting the open sets.

Depends on

Used by

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

Dependency tree · two levels

69 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

Dense set (Wikipedia) (standard reference, not scraped)
Trivial topology (Wikipedia) (standard reference, not scraped)
Hausdorff space (Wikipedia) (standard reference, not scraped)
General Topology Notes (UC Riverside) (standard reference, not scraped)
Cocountable topology (Wikipedia) (standard reference, not scraped)