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

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Two continuous maps $\mathbb{R} \to \mathbb{R}$ agreeing at every rational are equal

Example

Let $f, g : \mathbb{R} \to \mathbb{R}$ be continuous (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point) and suppose

$f(q) = g(q) \qquad \text{for every } q \in \mathbb{Q}_{\mathbb{R}} ,$

where $\mathbb{Q}_{\mathbb{R}}$ is the set of rationals inside $\mathbb{R}$ (The rationals embed densely in the reals). Then $f = g$ .

So a continuous real function is determined by its values at the rationals, and two continuous functions that are visibly different must already differ at some rational. Nothing is claimed about which functions on $\mathbb{Q}_{\mathbb{R}}$ extend continuously to $\mathbb{R}$ ; the statement is about uniqueness of the extension only.

Facts & Assumptions

Given: Continuous $f, g : \mathbb{R} \to \mathbb{R}$ agreeing at every rational, with $\mathbb{R}$ carrying its usual topology.

[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]

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

[L1]

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

[L2]

$\mathbb{R}$ with its usual topology is Hausdorff, being metrizable (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[L3]

For a real function, $\varepsilon$ - $\delta$ continuity is continuity as a map of topological spaces (Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$ , continuity and uniform continuity of $f : A \to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace, claim 1, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Continuity of a map of topological spaces at a point and globally).

[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).

Verification

technique · direct

1.1

$\mathbb{Q}_{\mathbb{R}}$ 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 [L1] some rational lies strictly between $x - r$ and $x + r$ , hence in $U$ .

A1A2L1

1.2

$\mathbb{R}$ is Hausdorff and both $f$ and $g$ are continuous as maps of topological spaces.

L2L3

2.1

By [L4] applied with domain $\mathbb{R}$ , dense subset $\mathbb{Q}_{\mathbb{R}}$ and Hausdorff codomain $\mathbb{R}$ , the hypothesis $f = g$ on $\mathbb{Q}_{\mathbb{R}}$ gives $f = g$ .

step 1.1step 1.2L4∎

Remarks

Why the density is established from the order rather than quoted. What Two continuous maps into a Hausdorff space that agree on a dense subset are equal needs is density in the sense of Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets, a condition on the open sets of $\mathbb{R}$ ; step 1.1 derives exactly that condition from the statement that a rational lies strictly between any two reals (The rationals embed densely in the reals), which is where the order of $\mathbb{R}$ enters and the only place it does.
Both hypotheses on the codomain and on the maps are needed. Dropping continuity of one map leaves the conclusion false for the obvious reason, and dropping the Hausdorff condition on the codomain leaves it false for a less obvious one: FALSE: two continuous maps that agree on a dense subset of their common domain are equal, with no hypothesis on the codomain exhibits two continuous maps on $\mathbb{R}$ agreeing at every rational, differing at every irrational, and taking values in a two-point space that is not Hausdorff.
The rationals enter at one step only. They are used in step 1.1 and nowhere else; the argument as written proves the statement for an arbitrary dense subset of $\mathbb{R}$ once density in the sense of Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets is in hand, and step 1.1 is simply where that is established for $\mathbb{Q}_{\mathbb{R}}$ .

Depends on

Used by

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Sources

Dense set (Wikipedia) (standard reference, not scraped)
Continuous function (Wikipedia) (standard reference, not scraped)
General Topology Notes (UC Riverside) (standard reference, not scraped)

Two continuous maps R→R\mathbb{R} \to \mathbb{R}R→R agreeing at every rational are equal