CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck 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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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 $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

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

$\bigcap_{k \ge 1}\Big(0, \frac{1}{k}\Big) = \emptyset .$

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 $j \in \mathbb{N}$ the open interval $J_j := \{x \in \mathbb{R} : 0 < x < 1/(j+1)\}$ , which is the family $(0,1/k)$ for $k \ge 1$ 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 $\mathbb{R}$ : 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 $\varepsilon > 0$ 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]

Trichotomy of the order on $\mathbb{R}$ (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 $j \in \mathbb{N}$ .

step 1.1L1

3.1

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

step 2.1L2L3L4

4.1

So no such $x$ exists: $\bigcap_j J_j = \emptyset$ , 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 $\mathbb{R}$ , supplied by For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , 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

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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)(0,1/k) have empty intersection