ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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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.

Every nondegenerate closed interval is perfect, giving a second proof that it is uncountable

Example

Let $a, b \in R$ with $a < b$ . Then the closed interval $E : = [a, b]$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) is perfect (Perfect subset of $R$ : closed with no isolated points), and therefore uncountable by Every nonempty perfect subset of $R$ is uncountable.

This is a second proof of the uncountability of a nondegenerate interval. The first, Every nondegenerate interval of $R$ is uncountable, runs a trisection argument directly against an assumed enumeration; the route here checks two purely local properties, closedness and the absence of isolated points, and lets the perfect-set theorem do the counting.

Facts & Assumptions

Given: Reals $a < b$ and the interval $E : = [a, b] = {x : a \leq x \leq b}$ .

[L1]

A set is perfect when it is closed and no point of it is isolated in it; $x \in P$ is isolated in $P$ when some $N_{ε} (x)$ meets $P$ only in $x$ (Perfect subset of $R$ : closed with no isolated points, Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ).

[L2]

Each interval of the form $[a, b]$ is a closed set (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L3]

$N_{ε} (x) = {y : ∣ y - x ∣ < ε}$ (The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ , Basic properties of the absolute value).

[L4]

Every nonempty perfect subset of $R$ is uncountable (Every nonempty perfect subset of $R$ is uncountable).

[L5]

For $a < b$ the intervals $[a, b]$ and $(a, b)$ are uncountable (Every nondegenerate interval of $R$ is uncountable).

[L6]

Every nonempty finite set of reals has a minimum, which is one of its members (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L7]

Ordered-field arithmetic: $0 < 1$ , so $2 : = 1 + 1 > 0$ and $0 < d \cdot 2^{- 1} < d$ for $d > 0$ ; adding a constant and multiplying by a positive preserve inequalities; the order is total and transitive (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Verification

technique · direct

1.1

$E$ is closed by [L2], and nonempty since $a \in E$ .

1.2

No point of $E$ is isolated in $E$ : let $x \in E$ and let $ε > 0$ be real. If $x < b$ , put $t : = \min {ε, b - x} \cdot 2^{- 1}$ , which is positive by [L6] and [L7], and $y : = x + t$ ; then $y > x$ , and $y \leq x + (b - x) \cdot 2^{- 1} < b$ by [L7], while $y > x \geq a$ , so $y \in E$ with $y \neq x$ and $∣ y - x ∣ = t \leq ε \cdot 2^{- 1} < ε$ . If $x = b$ , then $x > a$ ; put $t : = \min {ε, b - a} \cdot 2^{- 1} > 0$ and $y : = x - t$ ; then $y < x \leq b$ , and $y \geq b - (b - a) \cdot 2^{- 1} > a$ by [L7], so $y \in E$ with $y \neq x$ and $∣ y - x ∣ = t < ε$ . In both cases $N_{ε} (x)$ contains a point of $E$ other than $x$ , so no $ε$ isolates $x$ .

L1L3L6L7

2.1

By steps 1.1 and 1.2 the set $E$ is closed with no isolated points, that is, perfect, and it is nonempty.

step 1.1step 1.2L1

3.1

By [L4] the nonempty perfect set $E$ is uncountable, which reproves for $[a, b]$ the first claim of [L5] along an independent route.

step 1.1step 2.1L4L5∎

Remarks

Nondegeneracy is exactly what is needed. For $a = b$ the set $[a, a] = {a}$ is closed, its single point is isolated, and it is finite; the argument of step 1.2 breaks precisely there, since neither $x < b$ nor $x > a$ holds. This matches the hypothesis $a < b$ of Every nondegenerate interval of $R$ is uncountable.
The open interval supplies only half of the definition. The computation of step 1.2 applies verbatim inside $(a, b)$ and shows it has no isolated points, but $(a, b)$ is not closed, so it is not perfect. Perfectness needs both halves, which is why the example is stated for the closed interval.
Two proofs of one fact, sharing one ingredient. Both routes spend the completeness of $R$ exactly once, Every nondegenerate interval of $R$ is uncountable as a supremum and Every nonempty perfect subset of $R$ is uncountable 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 $0$ . They differ in everything else: the first trisects a given interval against a given enumeration, the second selects rational-endpoint intervals by least index. Neither is a corollary of the other.

Depends on

Used by

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Sources

Perfect set (Wikipedia) (standard reference, not scraped)
Interval (mathematics) (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (Thm 2.43 and its corollary) (standard reference, not scraped)