CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passverified 2026-08-05 (claude-sonnet-5)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The nested open intervals $(0, 1 / k)$ have empty intersection

Statement refuted

Refuted claim: a nested sequence of nonempty bounded open intervals has nonempty intersection (FALSE: a nested sequence of nonempty bounded open intervals has nonempty intersection, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

The witness is $J_{k} = (0, 1 / k)$ for $k \geq 1$ : each is a nonempty bounded open interval, the family is nested, and

$⋂_{k \geq 1} (0, \frac{1}{k}) = \emptyset .$

The refutation is carried out in full in FALSE: a nested sequence of nonempty bounded open intervals has nonempty intersection and is recorded here as the named counterexample. The comparison worth keeping in view is the closed family $[0, 1 / k]$ , which differs only by the inclusion of the left endpoint and intersects in ${0}$ ; that computation is The nested intervals $[0, 1 / k]$ intersect in exactly ${0}$ .

Facts & Assumptions

Given: For $j \in N$ the open interval $J_{j} : = {x \in R : 0 < x < 1 / (j + 1)}$ , which is the family $(0, 1 / k)$ for $k \geq 1$ under the substitution $k = j + 1$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

The family $(J_{j})$ consists of nonempty bounded open intervals, is nested, and has empty intersection (FALSE: a nested sequence of nonempty bounded open intervals has nonempty intersection, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

Canonical naturals are positive and strictly increasing in the index (Canonical naturals are positive and strictly increasing); reciprocals of positives are positive and reciprocation reverses the order (Inverses of positives are positive, and reciprocation reverses order).

[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]

Trichotomy of the order on $R$ (Complete ordered field (least-upper-bound property), Ordered field).

[L5]

The refuted claim: a nested sequence of nonempty bounded open intervals has nonempty intersection.

Counterexample

technique · direct

1.1

Each $J_{j}$ is a nonempty bounded open interval and $J_{j + 1} \subseteq J_{j}$ , so the family is an instance of the claim, which asserts that its intersection is nonempty.

givenL1L5

2.1

Suppose $x$ belonged to every $J_{j}$ . Then $x > 0$ , and $x < 1 / (j + 1)$ for every $j \in N$ .

step 1.1L1

3.1

Since $x > 0$ , fix a natural $n \geq 1$ with $1 / n < x$ , and write $n = j + 1$ with $j \in N$ ; step 2.1 then gives $x < 1 / n$ as well, which trichotomy forbids.

step 2.1L2L3L4

4.1

So no such $x$ exists: $⋂_{j} J_{j} = \emptyset$ , and the claim is refuted by a family of nonempty bounded open intervals.

step 1.1step 3.1L1L5∎

Remarks

Exactly one hypothesis of A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ is missing. The intervals here are nonempty, bounded and nested; they are not closed. The true theorem is therefore not contradicted, and the counterexample shows that its closedness hypothesis cannot be dropped.
The candidate point exists and is excluded by a hair. With $J_{j} = (a_{j}, b_{j})$ one still has $a_{j} \to 0$ and $b_{j} \to 0$ , and the only possible common point is $0$ ; but $0 \notin J_{j}$ for any $j$ , because the left endpoint is excluded. Closedness is precisely the hypothesis that puts the limiting endpoint into each set. Compare The nested intervals $[0, 1 / k]$ intersect in exactly ${0}$ , where it is present and the intersection is ${0}$ .
The Archimedean property is doing the work in step 3.1. In a non-Archimedean ordered field a positive infinitesimal lies in every $(0, 1 / n)$ , and the intersection is nonempty. So this is a counterexample about $R$ , supplied by For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , and not a formal consequence of openness alone.
Boundedness is a separate hypothesis and fails separately. The nested closed unbounded sets $[k, \infty)$ have empty intersection, so boundedness cannot be dropped keeps closedness and drops boundedness, with the same empty intersection, so neither hypothesis implies the other.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Nested intervals (Wikipedia) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)

The nested open intervals (0,1/k)(0,1/k) have empty intersection