CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
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$\mathbb{Q}$ is $F_\sigma$ , meager and not $G_\delta$ , while the irrationals are $G_\delta$ , residual and not $F_\sigma$

Statement

Write $\mathbb{Q}_{\mathbb{R}}$ for the image of $\mathbb{Q}$ in $\mathbb{R}$ under the canonical embedding $q \mapsto \hat q$ (The rationals embed densely in the reals), the set usually written $\mathbb{Q}$ once the identification is made, and put $X := \mathbb{R} \setminus \mathbb{Q}_{\mathbb{R}}$ for the irrationals. Then:

$\mathbb{Q}_{\mathbb{R}}$ is an $F_\sigma$ set ( $F_\sigma$ and $G_\delta$ subsets of $\mathbb{R}$ ) and is meager (Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$ );
$X$ is a $G_\delta$ set and is residual;
$\mathbb{Q}_{\mathbb{R}}$ is not a $G_\delta$ set, and $X$ is not an $F_\sigma$ set.

Claims 1 and 2 are bookkeeping. Claim 3 is the substance and is exactly where 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 is spent: no argument from the algebra of open and closed sets alone can reach it, since $\mathbb{Q}_{\mathbb{R}}$ and $X$ are interchanged by complementation while $F_\sigma$ and $G_\delta$ are, so any such argument would prove the same thing about both sets and about neither.

Facts & Assumptions

Given: The complete ordered field $\mathbb{R}$ , the set $\mathbb{Q}_{\mathbb{R}} \subseteq \mathbb{R}$ of rationals and its complement $X = \mathbb{R} \setminus \mathbb{Q}_{\mathbb{R}}$ .

[L1]

$\mathbb{Q} \approx \mathbb{N}$ ( $\mathbb{Q}$ is countably infinite, Equinumerous sets, $A \approx B$ and $A \preceq B$ ), $q \mapsto \hat q$ is injective with image $\mathbb{Q}_{\mathbb{R}}$ (The rationals embed densely in the reals), and a composition of bijections is a bijection (Injection, surjection, bijection).

[L2]

$\mathbb{Q}_{\mathbb{R}}$ is dense in $\mathbb{R}$ (Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$ , and every nonempty open subset of $\mathbb{R}$ is uncountable); a set is dense when its closure is $\mathbb{R}$ , equivalently when every $N_\varepsilon(x)$ meets it (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ , The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ); the closure operator is monotone, so a superset of a dense set is dense (Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ ).

[L3]

$U$ is open when every point of it has a neighbourhood inside it, and $F$ is closed when $\mathbb{R} \setminus F$ is open; $N_\varepsilon(x) = (x - \varepsilon, x + \varepsilon)$ and $x \in N_\varepsilon(x)$ (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L4]

A closed set is nowhere dense exactly when its interior is empty; a meager set is a union of a sequence of nowhere dense sets; residual means the complement is meager (Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$ , Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ , The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points).

[L5]

$A$ is $F_\sigma$ when it is the union of a sequence of closed sets and $G_\delta$ when it is the intersection of a sequence of open sets; $A$ is $F_\sigma$ if and only if $\mathbb{R} \setminus A$ is $G_\delta$ ( $F_\sigma$ and $G_\delta$ subsets of $\mathbb{R}$ ).

[L6]

A countable intersection of dense open subsets of $\mathbb{R}$ is dense (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).

[L7]

There is a bijection $J : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ ( $\mathbb{N} \times \mathbb{N} \approx \mathbb{N}$ ).

Proof

technique · contradiction

1.1

For $c \in \mathbb{R}$ the singleton $\{c\}$ is closed and nowhere dense: its complement is open, since $x \ne c$ gives $N_{|x-c|}(x) \subseteq \mathbb{R} \setminus \{c\}$ by [L3]; and its interior is empty, since for every real $\varepsilon > 0$ the point $c + \varepsilon \cdot 2^{-1}$ lies in $N_\varepsilon(c)$ and differs from $c$ , so no neighbourhood is contained in $\{c\}$ , whence $\{c\}$ is a closed set with empty interior and [L4] applies.

L3L4

1.2

By [L1] fix a bijection $\beta : \mathbb{N} \to \mathbb{Q}$ and put $e := \iota \circ \beta$ with $\iota(q) = \hat q$ , a bijection from $\mathbb{N}$ onto $\mathbb{Q}_{\mathbb{R}}$ .

L1choose

2.1

$\mathbb{Q}_{\mathbb{R}} = \bigcup_{n \in \mathbb{N}} \{e(n)\}$ , since $e$ is onto $\mathbb{Q}_{\mathbb{R}}$ ; the sets $\{e(n)\}$ are closed and nowhere dense by step 1.1, so $\mathbb{Q}_{\mathbb{R}}$ is $F_\sigma$ by [L5] and meager by [L4]. This is claim 1.

step 1.1step 1.2L4L5

3.1

Put $W_n := \mathbb{R} \setminus \{e(n)\}$ , an open set by step 1.1 and [L3]. A real $x$ lies in $\bigcap_n W_n$ exactly when $x \ne e(n)$ for every $n$ , that is, exactly when $x \notin \mathbb{Q}_{\mathbb{R}}$ , so $X = \bigcap_n W_n$ and $X$ is $G_\delta$ by [L5]; and $\mathbb{R} \setminus X = \mathbb{Q}_{\mathbb{R}}$ is meager by step 2.1, so $X$ is residual by [L4]. This is claim 2. Each $W_n$ is also dense, since every $N_\varepsilon(x)$ contains two distinct points and so meets $\mathbb{R} \setminus \{e(n)\}$ , by [L2] and [L3].

step 1.1step 1.2step 2.1L2L3L4L5

4.1

Suppose, for contradiction, that $\mathbb{Q}_{\mathbb{R}}$ is $G_\delta$ , and by [L5] fix a sequence $(V_n)$ of open sets with $\mathbb{Q}_{\mathbb{R}} = \bigcap_n V_n$ . Each $V_n$ contains $\mathbb{Q}_{\mathbb{R}}$ , which is dense by [L2], so each $V_n$ is dense by [L2]; and each $W_n$ of step 3.1 is open and dense.

assume-contrastep 3.1L2L5choose

5.1

By [L7] fix a bijection $J : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ and define a sequence $(D_j)$ by $D_{J(m,n)} := V_n$ when $m = 0$ and $D_{J(m,n)} := W_n$ when $m \ne 0$ ; this is total because $J$ is a bijection, and every $D_j$ is open and dense by step 4.1. Moreover $\bigcap_j D_j = \big(\bigcap_n V_n\big) \cap \big(\bigcap_n W_n\big) = \mathbb{Q}_{\mathbb{R}} \cap X = \varnothing$ , since every $V_n$ and every $W_n$ occurs among the $D_j$ and every $D_j$ is one of them.

step 3.1step 4.1L7

6.1

By [L6] the set $\bigcap_j D_j$ is dense, hence nonempty by [L2] and [L3], contradicting step 5.1. The assumption of step 4.1 is therefore untenable: $\mathbb{Q}_{\mathbb{R}}$ is not $G_\delta$ ; and $X$ is not $F_\sigma$ , since $\mathbb{R} \setminus X = \mathbb{Q}_{\mathbb{R}}$ would then be $G_\delta$ by [L5]. This is claim 3.

step 4.1step 5.1L2L3L5L6discharge-contradiction∎

Remarks

Where the two halves of the argument part company. Claim 1 is a listing argument: step 1.2 lists $\mathbb{Q}_{\mathbb{R}}$ , and step 1.1 shows that each real singleton is nowhere dense; claim 3 uses the completeness of $\mathbb{R}$ 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. Indeed $\mathbb{Q}$ is a $G_\delta$ subset of itself, being the whole space, so no argument that ignores the ambient completeness can possibly give claim 3.
The irrationals are also uncountable (The irrationals are uncountable), by a different and much cheaper argument that needs only the countability of $\mathbb{Q}$ and the uncountability of $\mathbb{R}$ . Uncountability and being residual are independent properties: $\mathbb{Q}$ is meager and countable, the Cantor set is meager and uncountable (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points), and $X$ is residual and uncountable.
The named witness for claim 3 is The irrationals form a residual $G_\delta$ set that is not $F_\sigma$ ↗, and the false statement it refutes is FALSE: $\mathbb{Q}$ is a $G_\delta$ subset of $\mathbb{R}$ ; the refutation is carried out here.
Meagre and measure zero are not the same smallness. $\mathbb{Q}_{\mathbb{R}}$ is both, but the two notions diverge as soon as one leaves the countable case: $\mathbb{R}$ is the union of a meager set and a set of measure zero, so smallness of category and smallness of measure are independent notions ↗ writes $\mathbb{R}$ as a meager set together with a set of measure zero, and the set of measure zero there is residual. So being residual, which is what claim 2 gives for $X$ , carries no information at all about size in measure.

Depends on

Used by

No function ℝ → ℝ is continuous at every rational and discontinuous at every irrational, because ℚ is not G_δ Corollary
The irrationals form a residual G_δ set that is not F_σ Counterexample
FALSE: ℚ is a G_δ subset of ℝ False statement

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 121 results over 37 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)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (Exercise 22) (standard reference, not scraped)
E. Zakon, Problems on Baire Categories and Linear Maps (standard reference, not scraped)
E. Zakon, Mathematical Analysis, §6.8: Baire Categories (standard reference, not scraped)

Q\mathbb{Q}Q is FσF_\sigmaFσ​, meager and not GδG_\deltaGδ​, while the irrationals are GδG_\deltaGδ​, residual and not FσF_\sigmaFσ​