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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$\mathbb{Q}$ has closure $\mathbb{R}$ , empty interior, and boundary $\mathbb{R}$

Example

Write $\mathbb{Q}_{\mathbb{R}}$ for the copy of $\mathbb{Q}$ inside $\mathbb{R}$ (The rationals embed densely in the reals). Then

$\overline{\mathbb{Q}_{\mathbb{R}}} = \mathbb{R}, \qquad (\mathbb{Q}_{\mathbb{R}})^{\circ} = \varnothing, \qquad \partial \mathbb{Q}_{\mathbb{R}} = \mathbb{R},$

with closure, interior and boundary as in Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ . So the rationals are as large as possible for the closure operator and as small as possible for the interior operator at once, and their boundary is everything.

Facts & Assumptions

Given: The copy $\mathbb{Q}_{\mathbb{R}}$ of $\mathbb{Q}$ in $\mathbb{R}$ and the set $X := \mathbb{R} \setminus \mathbb{Q}_{\mathbb{R}}$ of irrationals.

[L1]

Interior, closure and boundary: $\partial A = \overline{A} \setminus A^{\circ}$ , and $x \in A^{\circ}$ exactly when some $N_\varepsilon(x)$ is contained in $A$ (Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ ).

[L2]

Both $\mathbb{Q}_{\mathbb{R}}$ and $X$ are dense in $\mathbb{R}$ , that is, each has closure $\mathbb{R}$ (Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$ , and every nonempty open subset of $\mathbb{R}$ is uncountable).

[L3]

$\overline{A}$ is exactly the set of points every neighbourhood of which meets $A$ (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 $\mathbb{R}$ ).

[L4]

$N_\varepsilon(x) = (x-\varepsilon, x+\varepsilon)$ (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).

Verification

technique · direct

1.1

$\overline{\mathbb{Q}_{\mathbb{R}}} = \mathbb{R}$ : this is the density of $\mathbb{Q}_{\mathbb{R}}$ in [L2].

1.2

$(\mathbb{Q}_{\mathbb{R}})^{\circ} = \varnothing$ : suppose $x$ were in the interior; by [L1] there would be a real $\varepsilon > 0$ with $N_\varepsilon(x) \subseteq \mathbb{Q}_{\mathbb{R}}$ . But $\overline{X} = \mathbb{R}$ by [L2], so $x \in \overline{X}$ and every neighbourhood of $x$ meets $X$ by [L3]; a point of $N_\varepsilon(x) \cap X$ then lies in $\mathbb{Q}_{\mathbb{R}}$ and in its complement at once, which is impossible.

L1L2L3L4

1.3

By [L1] the boundary is $\partial \mathbb{Q}_{\mathbb{R}} = \overline{\mathbb{Q}_{\mathbb{R}}} \setminus (\mathbb{Q}_{\mathbb{R}})^{\circ}$ .

2.1

Substituting steps 1.1 and 1.2 into step 1.3 gives $\partial \mathbb{Q}_{\mathbb{R}} = \mathbb{R} \setminus \varnothing = \mathbb{R}$ , so all three assertions hold.

step 1.1step 1.2step 1.3L1∎

Remarks

The same computation applies verbatim to the irrationals. $X$ is dense by [L2] and its complement $\mathbb{Q}_{\mathbb{R}}$ is dense too, so $\overline{X} = \mathbb{R}$ , $X^{\circ} = \varnothing$ and $\partial X = \mathbb{R}$ . Two complementary sets can therefore both have boundary everything, which is what the density of each of them forces.
Empty interior is not smallness in any counting sense. $X$ has empty interior and is uncountable (The irrationals are uncountable), while $\mathbb{Q}_{\mathbb{R}}$ has empty interior and is countable ( $\mathbb{Q}$ is countably infinite). The interior measures whether the set contains an interval, and nothing else.
Where the density of the irrationals comes from. It is proved in Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$ , and every nonempty open subset of $\mathbb{R}$ is uncountable by counting: an interval is uncountable and the rationals are not, so an interval cannot consist of rationals alone. No explicit irrational is needed for this computation, though one is available (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).

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 77 results over 26 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)
Boundary (topology) (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)
J. K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

Q\mathbb{Q}Q has closure R\mathbb{R}R, empty interior, and boundary R\mathbb{R}R