CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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The nested closed unbounded sets $[k, \infty)$ have empty intersection, so boundedness cannot be dropped

Statement refuted

Refuted claim: a nested sequence of nonempty closed intervals has nonempty intersection, boundedness being unnecessary (A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

The witness is $I_{k} = [k, \infty)$ for $k \in N$ , where $k$ denotes the canonical natural of $R$ . Each is a nonempty closed interval, the family is nested, and

$⋂_{k \in N} [k, \infty) = \emptyset .$

A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ therefore cannot be improved by deleting "bounded" from its hypotheses. Together with the open-interval counterexample on this page, which deletes "closed" instead, this shows that the two hypotheses are independent and that neither is an artefact of the proof.

Facts & Assumptions

Given: For $k \in N$ the set $I_{k} : = {x \in R : k \leq x}$ , where $k$ denotes the canonical natural $k \cdot 1_{R}$ ; this is the closed interval $[k, \infty)$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

Intervals: $[a, \infty) = {x : a \leq x}$ is a closed interval, it is not bounded above, and it is nonempty since it contains $a$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

Canonical naturals: $m \mapsto m \cdot 1_{R}$ is strictly increasing, so $k \leq k + 1$ in $R$ for every $k \in N$ (Canonical naturals are positive and strictly increasing).

[L3]

Archimedean property: for every real $x$ there is a natural $n \geq 1$ with $x < n \cdot 1_{R}$ (Every complete ordered field is Archimedean).

[L4]

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

[L5]

Nested interval property, for nonempty closed bounded intervals (A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ ).

[L6]

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

Counterexample

technique · direct

1.1

Each $I_{k}$ is a nonempty closed interval, containing the canonical natural $k$ ; and none of them is bounded, since $[a, \infty)$ has no upper bound.

givenL1

2.1

The family is nested: $k \leq k + 1$ in $R$ , so $k + 1 \leq x$ implies $k \leq x$ , that is $I_{k + 1} \subseteq I_{k}$ .

step 1.1L2L4

3.1

Suppose $x$ belonged to every $I_{k}$ . Then $k \cdot 1_{R} \leq x$ for every $k \in N$ .

step 2.1L1

4.1

By the Archimedean property fix a natural $n \geq 1$ with $x < n \cdot 1_{R}$ ; step 3.1 applied to $k = n$ gives $n \cdot 1_{R} \leq x$ , which trichotomy forbids.

step 3.1L3L4

5.1

So no such $x$ exists: the family $(I_{k})$ consists of nonempty closed intervals, is nested, and has empty intersection. The claim is refuted, and boundedness cannot be dropped from A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ .

step 1.1step 2.1step 4.1L5L6∎

Remarks

What fails is the existence of the supremum, not the argument's bookkeeping. In the proof 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 $0$ the intersection is computed as $[\sup a_{k}, \inf b_{k}]$ . Here $a_{k} = k$ and the set ${k : k \in N}$ has no supremum in $R$ , precisely because $R$ is Archimedean, so there is no candidate point at all. This is a different failure mode from The nested open intervals $(0, 1 / k)$ have empty intersection, where the candidate point exists and is merely not a member.
The sets are intervals but not bounded intervals. Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length admits $[a, \infty)$ as one of its nine forms and assigns it no length, which is exactly why the length hypothesis of the nested interval property has nothing to say about this family.
The Archimedean property is again what makes the intersection empty. In a non-Archimedean ordered field an element exceeding every canonical natural lies in every $I_{k}$ , so the intersection is nonempty there. The counterexample is a statement about $R$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 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

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 closed unbounded sets [k,∞)[k,∞) have empty intersection, so boundedness cannot be dropped