LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

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Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$ , and every nonempty open subset of $\mathbb{R}$ is uncountable

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 dense in $\mathbb{R}$ , that is, $\overline{\mathbb{Q}_{\mathbb{R}}} = \mathbb{R}$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ );
$X$ is dense in $\mathbb{R}$ ;
every nonempty open subset of $\mathbb{R}$ is uncountable (Finite, countably infinite, countable, uncountable).

Claim 2 is not a symmetry of claim 1: the rationals are dense because they are constructed to approximate, whereas the irrationals are dense because there are too many points in any interval for a countable set to exhaust it, which is why claim 3 is proved alongside and used for it.

Facts & Assumptions

Given: The canonical embedding $q \mapsto \hat q$ of $\mathbb{Q}$ into $\mathbb{R}$ , its image $\mathbb{Q}_{\mathbb{R}}$ , and the complement $X = \mathbb{R} \setminus \mathbb{Q}_{\mathbb{R}}$ .

[L1]

$\overline{A}$ is the set of points every neighbourhood of which meets $A$ ; $A$ is dense in $\mathbb{R}$ when $\overline{A} = \mathbb{R}$ (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, Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ).

[L2]

$N_\varepsilon(x) = (x - \varepsilon, x + \varepsilon)$ for real $\varepsilon > 0$ , and $x - \varepsilon < x + \varepsilon$ (The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ , Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L3]

$U$ is open when every $x \in U$ admits $\varepsilon > 0$ with $N_\varepsilon(x) \subseteq U$ (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L4]

Strictly between any two reals lies an element of $\mathbb{Q}_{\mathbb{R}}$ , and $q \mapsto \hat q$ is injective (The rationals embed densely in the reals).

[L5]

$\mathbb{Q} \approx \mathbb{N}$ ( $\mathbb{Q}$ is countably infinite); an injection is a bijection onto its image, and $\approx$ is symmetric and transitive (Injection, surjection, bijection, Equinumerous sets, $A \approx B$ and $A \preceq B$ ).

[L6]

Every subset of an at most countable set is at most countable, and uncountable means not at most countable (Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable).

[L7]

For $a < b$ the interval $(a,b)$ is uncountable (Every nondegenerate interval of $\mathbb{R}$ is uncountable).

Proof

technique · direct

1.1

$\mathbb{Q}_{\mathbb{R}}$ is dense: let $x \in \mathbb{R}$ and let $\varepsilon > 0$ be real; by [L2] one has $x - \varepsilon < x + \varepsilon$ , so [L4] supplies $\hat q$ with $x - \varepsilon < \hat q < x + \varepsilon$ , that is $\hat q \in N_\varepsilon(x) \cap \mathbb{Q}_{\mathbb{R}}$ . Every real is therefore an adherent point of $\mathbb{Q}_{\mathbb{R}}$ and claim 1 follows from [L1].

L1L2L4

1.2

$\mathbb{Q}_{\mathbb{R}}$ is at most countable: the embedding is an injection of $\mathbb{Q}$ with image $\mathbb{Q}_{\mathbb{R}}$ , hence a bijection onto it, so $\mathbb{Q}_{\mathbb{R}} \approx \mathbb{Q} \approx \mathbb{N}$ .

L4L5

1.3

For all reals $a < b$ the interval $(a,b)$ is uncountable.

2.1

For all reals $a < b$ the interval $(a,b)$ contains an irrational: if it did not, then $(a,b) \subseteq \mathbb{Q}_{\mathbb{R}}$ , so $(a,b)$ would be a subset of an at most countable set by step 1.2 and hence at most countable by [L6], contradicting step 1.3. So some $z \in (a,b)$ lies in $X$ .

step 1.2step 1.3L6

2.2

Every nonempty open $U \subseteq \mathbb{R}$ is uncountable: fix $x \in U$ and, by [L3], a real $\varepsilon > 0$ with $N_\varepsilon(x) \subseteq U$ ; by [L2] the set $N_\varepsilon(x)$ is the interval $(x - \varepsilon, x + \varepsilon)$ with $x - \varepsilon < x + \varepsilon$ , hence uncountable by step 1.3. Were $U$ at most countable, its subset $N_\varepsilon(x)$ would be at most countable by [L6], which it is not; so $U$ is uncountable, which is claim 3.

step 1.3L2L3L6choose

3.1

$X$ is dense: let $x \in \mathbb{R}$ and let $\varepsilon > 0$ be real; applying step 2.1 with $a = x - \varepsilon$ and $b = x + \varepsilon$ gives $z \in (x - \varepsilon, x + \varepsilon) \cap X$ , which is $N_\varepsilon(x) \cap X$ by [L2]. Every real is therefore an adherent point of $X$ , so $\overline{X} = \mathbb{R}$ by [L1], which is claim 2.

step 2.1L1L2

4.1

Claims 1, 2 and 3 are steps 1.1, 3.1 and 2.2, so both $\mathbb{Q}_{\mathbb{R}}$ and its complement are dense in $\mathbb{R}$ and every nonempty open subset of $\mathbb{R}$ is uncountable.

step 1.1step 2.2step 3.1∎

Remarks

Two dense sets can be disjoint. $\mathbb{Q}_{\mathbb{R}}$ and $X$ partition $\mathbb{R}$ and both are dense, so density says nothing about size: one of them is countable and the other is not (The irrationals are uncountable). What density does say is that neither has interior: a set whose complement is dense has empty interior, which is the computation carried out for $\mathbb{Q}$ in $\mathbb{Q}$ has closure $\mathbb{R}$ , empty interior, and boundary $\mathbb{R}$ ↗.
Claim 3 is a statement about open sets, not about intervals. It follows from the uncountability of intervals (Every nondegenerate interval of $\mathbb{R}$ is uncountable) only because openness supplies an interval inside the set at each of its points. A nonempty set with empty interior can perfectly well be countable, as $\mathbb{Q}_{\mathbb{R}}$ shows.
An explicit irrational is not produced here. Step 2.1 is a counting argument and exhibits nothing. The library does exhibit one separately, $\sqrt 2$ (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ , FALSE: some rational number squares to 2), and an explicit irrational in a given interval $(a,b)$ can be built from it as $q_1 + (q_2 - q_1)/\sqrt 2$ for suitable rationals $q_1 < q_2$ in the interval; that route is longer and is not the one taken above.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 107 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

Dense set (Wikipedia) (standard reference, not scraped)
Countable set (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 and Ch. 2 (standard reference, not scraped)
J. Lebl, Basic Analysis I, §1.2 (standard reference, not scraped)
J. K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

Both Q\mathbb{Q}Q and R∖Q\mathbb{R} \setminus \mathbb{Q}R∖Q are dense in R\mathbb{R}R, and every nonempty open subset of R\mathbb{R}R is uncountable