CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-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.

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 $\mathbb{R}$ (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

$\mathcal{U} \;:=\; \{\, V_k : k \ge 1 \,\}, \qquad V_k := (1/k,\ 1),$

of open subsets of $\mathbb{R}$ . The index runs over $k \ge 1$ because $1/0$ is undefined, and the first member is degenerate: $V_1 = (1,1) = \varnothing$ , 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 $k \ge 1$ the interval $V_k := (1/k,\ 1)$ , where $1/k$ is the inverse of the canonical natural $k \cdot 1_{\mathbb{R}}$ , and the family $\mathcal{U} := \{\, V_k : k \ge 1 \,\}$ .

[A1]

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

[L1]

An open cover of $S$ is a family of open subsets of $\mathbb{R}$ 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 $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset).

[L2]

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

[L3]

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

[L4]

Canonical naturals are positive and increasing for $k \ge 1$ , with $1 \cdot 1_{\mathbb{R}} = 1$ ; reciprocation of positives is positive and reverses the order, so $1 \le m \le n$ gives $0 < 1/n \le 1/m \le 1$ (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 $z \ge 0$ ; $0 < 1$ , 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 $\mathbb{R}$ is compact exactly when it is closed and bounded (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded).

Counterexample

technique · direct

1.1

Each $V_k$ is an open subset of $\mathbb{R}$ by [L2], and $\mathcal{U}$ covers $(0,1)$ : given $x$ with $0 < x < 1$ , [L3] supplies a natural $n \ge 1$ with $1/n < x$ , so $x \in V_n$ . Hence $\mathcal{U}$ 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 $\varepsilon > 0$ the point $t := \min\{\varepsilon, 1\} \cdot 2^{-1}$ is positive, satisfies $t \le 1 \cdot 2^{-1} < 1$ and $|t - 0| = t \le \varepsilon \cdot 2^{-1} < \varepsilon$ by [L5] and [L6], so $t \in N_\varepsilon(0) \cap (0,1)$ ; 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 $\mathcal{U}$ 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 $k_i \ge 1$ ; put $K := \max\{k_0, \dots, k_p\}$ by [L5], one of the $k_i$ , so $K \ge 1$ . Since $k_i \le K$ gives $1/K \le 1/k_i$ 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 $0 < z \le 1 \cdot 2^{-1} < 1$ by [L4] and [L6], since $K + 1 \ge 2$ , so $z \in (0,1)$ ; and $z \le 1/K$ by [L4], so $1/K < z$ fails and $z \notin V_K$ . Thus $z$ is uncovered.

step 1.2L1L4L5L6

3.1

The family $\mathcal{U}$ 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 $\mathcal{U}$ 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 $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ). 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 43 results over 14 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

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