CounterexampleConstruction: AI-generatedVerification: 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.

$\{0\} \cup [1,2]$ is closed, has an isolated point, and is not perfect

Statement refuted

Refuted claim: every closed subset of $\mathbb{R}$ is perfect (Perfect subset of $\mathbb{R}$ : closed with no isolated points).

The witness is $E := \{0\} \cup [1,2]$ . It is closed, being a union of two closed sets (Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets), and $0$ is an isolated point of it (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ), so the second clause of the definition of a perfect set fails while the first holds.

Facts & Assumptions

Given: The set $E := \{0\} \cup [1,2]$ , where $\{0\} = [0,0]$ and $[1,2]$ are closed bounded intervals (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[A1]

The refuted claim: every closed subset of $\mathbb{R}$ is perfect.

[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 $N_\varepsilon(x)$ satisfies $N_\varepsilon(x) \cap P = \{x\}$ (Perfect subset of $\mathbb{R}$ : closed with no isolated points, Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ).

[L2]

Each interval of the form $[a,b]$ is a closed set, and a union of finitely many closed sets is closed (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets).

[L3]

$N_\varepsilon(x) = \{\, y : |y - x| < \varepsilon \,\}$ , and $|y| = y$ for $y \ge 0$ (The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ , Basic properties of the absolute value).

[L4]

$0 < 1$ , so $2 := 1 + 1 > 0$ and $0 < 1 \cdot 2^{-1} < 1$ ; 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.

Counterexample

technique · direct

1.1

$E$ is closed: $\{0\} = [0,0]$ and $[1,2]$ are closed sets by [L2], and their union is closed by [L2]. So $E$ is a legitimate instance of the claim [A1].

A1L2

1.2

$0$ is an isolated point of $E$ : certainly $0 \in E$ ; put $\varepsilon := 1 \cdot 2^{-1}$ , which is positive and $< 1$ by [L4]. An element $y$ of $E$ is $0$ or lies in $[1,2]$ , and in the second case $|y - 0| = y \ge 1 > \varepsilon$ by [L3] and [L4], so $y \notin N_\varepsilon(0)$ . Hence $N_\varepsilon(0) \cap E = \{0\}$ .

L1L3L4

2.1

By step 1.2 the closed set $E$ of step 1.1 has an isolated point, so it is not perfect by [L1], and the claim [A1] is refuted.

step 1.1step 1.2A1L1∎

Remarks

Only the second clause fails, and by one point. Every point of $[1,2]$ is a limit point of $E$ , by the computation in Every nondegenerate closed interval is perfect, giving a second proof that it is uncountable; the single point $0$ is what stops $E$ from being perfect. Deleting it leaves $[1,2]$ , which is perfect.
Countability is the visible consequence. A nonempty perfect set is uncountable (Every nonempty perfect subset of $\mathbb{R}$ is uncountable). $E$ is uncountable too, since it contains $[1,2]$ , so this example does not separate the two notions by size; what it shows is that closedness alone does not give perfectness. A countable closed set with isolated points is $\{\, 1/k : k \ge 1 \,\} \cup \{0\}$ ( $\{1/k : k \ge 1\} \cup \{0\}$ is compact while $\{1/k : k \ge 1\}$ is not closed), and it is likewise not perfect.
The empty set is the degenerate case on the other side. It is closed and has no isolated points, hence is perfect, and it is countable; that is why Every nonempty perfect subset of $\mathbb{R}$ is uncountable assumes its perfect set is nonempty.

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: 28 results over 13 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

Perfect set (Wikipedia) (standard reference, not scraped)
Isolated point (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)

{0}∪[1,2]\{0\} \cup [1,2]{0}∪[1,2] is closed, has an isolated point, and is not perfect