False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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FALSE: $\mathbb{Q}$ is a $G_\delta$ subset of $\mathbb{R}$

Statement

False claim: $\mathbb{Q}$ , that is the set $\mathbb{Q}_{\mathbb{R}}$ of rationals inside $\mathbb{R}$ (The rationals embed densely in the reals), is a $G_\delta$ set ( $F_\sigma$ and $G_\delta$ subsets of $\mathbb{R}$ ): there is a sequence $(V_n)$ of open subsets of $\mathbb{R}$ with $\mathbb{Q}_{\mathbb{R}} = \bigcap_n V_n$ .

The claim looks plausible by symmetry. $\mathbb{Q}_{\mathbb{R}}$ is $F_\sigma$ , being a countable union of singletons; the irrationals are $G_\delta$ , being a countable intersection of complements of singletons; and the two classes are exchanged by complementation. So one expects each set to belong to both classes. It does not: the symmetry between the two classes says nothing about a single set, and the obstruction is the Baire category theorem.

Facts & Assumptions

Given: The set $\mathbb{Q}_{\mathbb{R}} \subseteq \mathbb{R}$ of rationals.

[A1]

The false claim: $\mathbb{Q}_{\mathbb{R}}$ is a $G_\delta$ subset of $\mathbb{R}$ .

[L1]

$\mathbb{Q}_{\mathbb{R}}$ is $F_\sigma$ and meager, the irrationals are $G_\delta$ and residual, and $\mathbb{Q}_{\mathbb{R}}$ is not $G_\delta$ ( $\mathbb{Q}$ is $F_\sigma$ , meager and not $G_\delta$ , while the irrationals are $G_\delta$ , residual and not $F_\sigma$ , claims 1, 2 and 3).

[L2]

$A$ is $G_\delta$ when it is the intersection of a sequence of open sets ( $F_\sigma$ and $G_\delta$ subsets of $\mathbb{R}$ , Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L3]

A countable intersection of dense open subsets of $\mathbb{R}$ is dense; in particular it is nonempty (Baire category in $\mathbb{R}$ , by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $\mathbb{R}$ is not a countable union of nowhere dense sets).

Refutation

technique · direct

1.1

By claim 3 of [L1], $\mathbb{Q}_{\mathbb{R}}$ is not a $G_\delta$ subset of $\mathbb{R}$ , which is the direct negation of [A1].

A1L1L2

1.2

The reason, recorded here so that the refutation is not merely a pointer: were $\mathbb{Q}_{\mathbb{R}} = \bigcap_n V_n$ with each $V_n$ open, every $V_n$ would contain the dense set $\mathbb{Q}_{\mathbb{R}}$ and so be dense; adjoining the dense open sets $\mathbb{R} \setminus \{q\}$ , one for each rational $q$ , would produce an at most countable family of dense open sets whose intersection is $\mathbb{Q}_{\mathbb{R}}$ minus every rational, that is $\varnothing$ , contradicting [L3].

L1L2L3

2.1

So [A1] is false, and the refutation is carried out in full in [L1].

step 1.1step 1.2A1∎

Remarks

What is true about $\mathbb{Q}$ . It is $F_\sigma$ , meager, of measure zero, dense, and countable. What fails is only the $G_\delta$ property, and its failure is a genuine theorem about $\mathbb{R}$ , resting on completeness through A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ inside Baire category in $\mathbb{R}$ , by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $\mathbb{R}$ is not a countable union of nowhere dense sets. Inside $\mathbb{Q}$ itself the corresponding claim is true and trivial, $\mathbb{Q}$ being the whole space there.
The dual false statement is not recorded separately, because it is the same statement: the irrationals fail to be $F_\sigma$ exactly because $\mathbb{Q}_{\mathbb{R}}$ fails to be $G_\delta$ ( $F_\sigma$ and $G_\delta$ subsets of $\mathbb{R}$ ). The witness is The irrationals form a residual $G_\delta$ set that is not $F_\sigma$ ↗.
Context, not a result of this library. In classical analysis the set of points at which a real function is continuous is always $G_\delta$ , and it is the false statement above that then rules out a function continuous at every rational and at no irrational. That classical result is not proved here, and continuity is not available at this point in the reading order; the connection is recorded as orientation and nothing on this page depends on it.

Depends on

Used by

The irrationals form a residual G_δ set that is not F_σ Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 97 results over 33 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

Gδ set (Wikipedia) (standard reference, not scraped)
Baire category theorem (Wikipedia) (standard reference, not scraped)
E. Zakon, Problems on Baire Categories and Linear Maps (standard reference, not scraped)

FALSE: Q\mathbb{Q}Q is a GδG_\deltaGδ​ subset of R\mathbb{R}R