CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck 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.

The cover ${(1 / k, 1)}$ of $(0, 1)$ has no finite subcover, so $(0, 1)$ is not compact

Statement refuted

Refuted claim: the bounded interval $(0, 1)$ is compact (Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset); equivalently, boundedness on its own is enough for compactness.

The witness is the family

$U : = {V_{k} : k \geq 1}, V_{k} : = (1 / k, 1),$

of open subsets of $R$ . The index runs over $k \geq 1$ because $1 / 0$ is undefined, and the first member is degenerate: $V_{1} = (1, 1) = \emptyset$ , which is harmless, since a cover may contain empty members. The family covers $(0, 1)$ and no finite subfamily does.

Facts & Assumptions

Given: For each natural $k \geq 1$ the interval $V_{k} : = (1 / k, 1)$ , where $1 / k$ is the inverse of the canonical natural $k \cdot 1_{R}$ , and the family $U : = {V_{k} : k \geq 1}$ .

[A1]

The refuted claim: $(0, 1)$ is compact.

[L1]

An open cover of $S$ is a family of open subsets of $R$ whose union contains $S$ ; $S$ is compact when every open cover has a finite subfamily, empty or of the form ${U_{0}, \dots, U_{p}}$ , whose union contains $S$ (Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset).

[L2]

Each interval of the form $(a, b)$ is an open set; $N_{ε} (x) = {y : ∣ y - x ∣ < ε}$ (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, The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L3]

Reciprocal Archimedean property: for every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean).

[L4]

Canonical naturals are positive and increasing for $k \geq 1$ , with $1 \cdot 1_{R} = 1$ ; reciprocation of positives is positive and reverses the order, so $1 \leq m \leq n$ gives $0 < 1 / n \leq 1 / m \leq 1$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order). 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.

[L5]

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

[L6]

Absolute value and ordered-field arithmetic: $∣ z ∣ = z$ for $z \geq 0$ ; $0 < 1$ , so $2 : = 1 + 1 > 0$ and $0 < d \cdot 2^{- 1} < d$ for $d > 0$ ; the order is total and transitive (Basic properties of the absolute value, 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.

[L7]

A subset of $R$ is compact exactly when it is closed and bounded (A subset of $R$ is compact if and only if it is closed and bounded).

Counterexample

technique · direct

1.1

Each $V_{k}$ is an open subset of $R$ by [L2], and $U$ covers $(0, 1)$ : given $x$ with $0 < x < 1$ , [L3] supplies a natural $n \geq 1$ with $1 / n < x$ , so $x \in V_{n}$ . Hence $U$ is an open cover of $(0, 1)$ .

L1L2L3L4

1.2

$(0, 1)$ is nonempty, since $1 \cdot 2^{- 1}$ satisfies $0 < 1 \cdot 2^{- 1} < 1$ by [L6].

1.3

$(0, 1)$ is not closed: $0 \notin (0, 1)$ , and for a real $ε > 0$ the point $t : = \min {ε, 1} \cdot 2^{- 1}$ is positive, satisfies $t \leq 1 \cdot 2^{- 1} < 1$ and $∣ t - 0 ∣ = t \leq ε \cdot 2^{- 1} < ε$ by [L5] and [L6], so $t \in N_{ε} (0) \cap (0, 1)$ ; hence no neighbourhood of $0$ lies in the complement of $(0, 1)$ and that complement is not open.

L2L5L6

2.1

No finite subfamily of $U$ covers $(0, 1)$ : the empty subfamily fails by step 1.2, and a nonempty finite subfamily is ${V_{k_{0}}, \dots, V_{k_{p}}}$ with each $k_{i} \geq 1$ ; put $K : = \max {k_{0}, \dots, k_{p}}$ by [L5], one of the $k_{i}$ , so $K \geq 1$ . Since $k_{i} \leq K$ gives $1 / K \leq 1 / k_{i}$ by [L4], each $V_{k_{i}}$ is contained in $V_{K} = (1 / K, 1)$ , so the union of the subfamily lies in $V_{K}$ . The point $z : = 1 / (K + 1)$ satisfies $0 < z \leq 1 \cdot 2^{- 1} < 1$ by [L4] and [L6], since $K + 1 \geq 2$ , so $z \in (0, 1)$ ; and $z \leq 1 / K$ by [L4], so $1 / K < z$ fails and $z \notin V_{K}$ . Thus $z$ is uncovered.

step 1.2L1L4L5L6

3.1

The family $U$ is therefore an open cover of $(0, 1)$ with no finite subcover, so $(0, 1)$ is not compact and the claim [A1] is refuted. This is consistent with [L7]: $(0, 1)$ is bounded but not closed by step 1.3, so [L7] predicts exactly this failure.

step 1.1step 1.3step 2.1A1L1L7∎

Remarks

The cover creeps up on the missing endpoint. Every member of $U$ stops short of $0$ , and the whole family reaches every point of $(0, 1)$ only because the reciprocals $1 / k$ get arbitrarily small (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ). A finite subfamily stops at the largest of its indices $K$ and therefore misses every point of $(0, 1)$ that is at most $1 / K$ .
The empty member is not a defect. $V_{1} = (1, 1)$ is empty because $1 / 1 = 1$ ; a family of open sets is a cover as long as its union contains the set, and an empty member contributes nothing either way. Writing the family from $k = 2$ instead would change nothing in the argument.
The closed interval behaves differently, and that is the whole point. $[0, 1]$ is compact by Heine-Borel by bisection: every closed bounded interval $[a, b]$ is compact, and the analogous family ${(1 / k, 1)}$ is not even a cover of it, since it misses both $0$ and $1$ . The endpoint that the cover above exploits is $0$ , which $(0, 1)$ omits and $[0, 1]$ contains.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

32 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Compact space (Wikipedia) (standard reference, not scraped)
Heine-Borel theorem (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (Example 2.21(h)) (standard reference, not scraped)
University of Colorado, Analysis I midterm solutions (standard reference, not scraped)

The cover {(1/k,1)}{(1/k,1)} of (0,1)(0,1) has no finite subcover, so (0,1)(0,1) is not compact