False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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FALSE: a nested sequence of nonempty bounded open intervals has nonempty intersection

Statement

False claim: if $(J_{k})_{k \in N}$ is a sequence of nonempty bounded open intervals of $R$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) with $J_{k + 1} \subseteq J_{k}$ for every $k$ , then $⋂_{k \in N} J_{k} \neq \emptyset$ .

The corresponding statement for closed bounded intervals is true and is 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 claim above is what one gets by replacing "closed" with "open" there, and it fails: the intersection can be empty. So closedness is not a convenience of the proof, it is a hypothesis without which the conclusion is false.

The witness is $J_{k} = (0, 1 / (k + 1))$ , refuted below and recorded separately as the named counterexample of the companion page. The index shift $1 / (k + 1)$ is the usual one for sequences starting at $k = 0$ ; in the customary notation the family is $(0, 1 / n)$ for $n \geq 1$ .

Facts & Assumptions

Given: For $k \in N$ the open interval $J_{k} : = {x \in R : 0 < x < 1 / (k + 1)}$ , where $k + 1$ denotes the canonical natural $(k + 1) \cdot 1_{R}$ , which is positive and invertible; this is a sequence of subsets of $R$ indexed by $N$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

Intervals: $(a, b) = {x : a < x < b}$ is an open interval, bounded, and nonempty whenever $a < b$ , since then $a < (a + b) / 2 < b$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

Canonical naturals: $n \cdot 1_{R} > 0$ for $n \geq 1$ , and $m \mapsto m \cdot 1_{R}$ is strictly increasing (Canonical naturals are positive and strictly increasing).

[L3]

Reciprocals: if $a > 0$ then $1 / a > 0$ , and $0 < a < b$ gives $0 < 1 / b < 1 / a$ (Inverses of positives are positive, and reciprocation reverses order).

[L4]

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).

[L5]

Trichotomy, so $x < y$ and $y < x$ cannot both hold (Complete ordered field (least-upper-bound property), Ordered field).

[L6]

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

Refutation

technique · direct

1.1

Each $J_{k}$ is an open interval and is bounded, with $0$ a lower bound and $1 / (k + 1)$ an upper bound.

givenL1

1.2

Each $J_{k}$ is nonempty: $k + 1 \geq 1$ gives $1 / (k + 1) > 0$ , so the endpoints satisfy $0 < 1 / (k + 1)$ and [L1] applies.

givenL1L2L3

2.1

The family is nested: $k + 1 < k + 2$ gives $0 < 1 / (k + 2) < 1 / (k + 1)$ , so $0 < x < 1 / (k + 2)$ implies $0 < x < 1 / (k + 1)$ , that is $J_{k + 1} \subseteq J_{k}$ .

step 1.2L2L3

2.2

So $(J_{k})$ is a sequence of nonempty bounded open intervals, nested, and is therefore an instance of the claim, which asserts that its intersection is nonempty.

step 1.1step 1.2L6

3.1

Suppose $x \in ⋂_{k} J_{k}$ . Then $x > 0$ , and $x < 1 / (k + 1)$ for every $k \in N$ .

step 2.2given

4.1

Since $x > 0$ , [L4] supplies a natural $n \geq 1$ with $1 / n < x$ ; writing $n = k + 1$ with $k \in N$ , which is possible because $n \geq 1$ , step 3.1 gives $x < 1 / n$ as well.

step 3.1L4choose

5.1

That is $x < 1 / n$ and $1 / n < x$ , which trichotomy forbids. So no such $x$ exists and $⋂_{k} J_{k} = \emptyset$ .

step 4.1L5

6.1

The sequence $(J_{k})$ therefore consists of nonempty bounded open intervals, is nested, and has empty intersection: the claim is false.

step 2.2step 5.1L6∎

Remarks

Which 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 being violated. Only closedness. The intervals here are nonempty and bounded, and the family is nested, so the true theorem does not apply, and the refutation shows that no weakening of it to open intervals is available.
What goes wrong in the proof of the true theorem. With $J_{k} = (a_{k}, b_{k})$ the endpoint sequences still converge, to $a = \sup a_{k}$ and $b = \inf b_{k}$ , and the intersection is still an interval with those endpoints; but for open intervals it is $[a, b]$ intersected with the open conditions, and here $a = 0 = b$ while $0 \notin J_{k}$ for any $k$ . The candidate point exists as a real number and simply fails to lie in the sets. Closedness is exactly the hypothesis that puts the endpoint into each interval.
The Archimedean property is what makes the intersection empty. In a non-Archimedean ordered field the same family has a nonempty intersection, since a positive infinitesimal lies below every $1 / n$ . So the counterexample is a statement about $R$ , and it is For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ that supplies it.
A closely related true statement. The intersection of the closures $[0, 1 / (k + 1)]$ is ${0}$ (The nested intervals $[0, 1 / k]$ intersect in exactly ${0}$ ↗), which is the same computation with the endpoint included, and it is exactly what the true theorem predicts once the lengths are seen to tend to $0$ .
The witness is recorded as the named counterexample The nested open intervals $(0, 1 / k)$ have empty intersection ↗.

Depends on

Used by

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

Dependency tree · two levels

27 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)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (Thm 2.38) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)