False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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FALSE: the nested interval property alone implies the least-upper-bound property

Statement

False claim: every ordered field with the nested interval property (NIP) of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness has the least-upper-bound property (LUB).

This is clause 2 of For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness with its Archimedean hypothesis deleted, and the deletion is exactly what makes it false. The witness is the formal Laurent series field $K = R ((t^{- 1}))$ , which satisfies (NIP) and has no least upper bound for the set of its own canonical naturals.

Note that the false claim is being refuted in the shrinking form of (NIP), which is the weaker hypothesis and therefore makes the implication stronger.

Facts & Assumptions

Given: The formal Laurent series field $K = R ((t^{- 1}))$ .

[L1]

$K$ is an ordered field ( $R ((t^{- 1}))$ is an ordered field, ordered by the sign of the leading coefficient).

[L2]

Every nested sequence of closed intervals of $K$ whose lengths tend to $0$ in $K$ has exactly one point in its intersection ( $R ((t^{- 1}))$ has the nested interval property for lengths tending to $0$ ); intervals, nesting and lengths tending to $0$ in an ordered field are as in Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field, and (NIP) asks exactly that such an intersection be nonempty (The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness).

[L3]

$K$ is not a complete ordered field: the set $A = {n \cdot 1_{K} : n \in N}$ is nonempty and bounded above by $t$ and has no least upper bound in $K$ ( $R ((t^{- 1}))$ does not have the least-upper-bound property; its canonical naturals have no supremum, Complete ordered field (least-upper-bound property)).

[L4]

$K$ is not Archimedean, since $n \cdot 1_{K} < t$ for every natural $n$ ( $R ((t^{- 1}))$ is non-Archimedean, and the monomials $t^{- k}$ are cofinal below its positive elements, Archimedean ordered field).

[L5]

For an ordered field, the Archimedean property together with (NIP) does imply (LUB) (For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness, clause 2 implies clause 1).

Refutation

technique · direct

1.1

$K$ is an ordered field.

1.2

$K$ has (NIP): any nested sequence of closed intervals of $K$ whose lengths tend to $0$ in $K$ has a point in its intersection, indeed exactly one.

1.3

$K$ does not have (LUB), the set of its canonical naturals being nonempty, bounded above and without a least upper bound.

2.1

So $K$ is an ordered field with (NIP) and without (LUB), and the claim is false.

step 1.1step 1.2step 1.3

3.1

What fails in $K$ is precisely the hypothesis that the claim deleted: $K$ is not Archimedean, and with that hypothesis restored the implication is true.

step 1.1L4L5∎

Remarks

The failure is not an accident of one field. By An ordered field with the least-upper-bound property has the nested interval property and is Archimedean every field with (LUB) is Archimedean, so any witness at all must be non-Archimedean; and in a non-Archimedean field the shrinking hypothesis in (NIP) is a severe restriction, because a length that tends to $0$ in the order of the field must get below every infinitesimal. That is why checking shrinking (NIP) in $K$ is substantive, and why $K$ can satisfy (NIP) while failing (LUB) at all.
$R (t)$ will not do as a witness, although it is the library's other non-Archimedean ordered field (Not every ordered field is Archimedean). Nothing in this library establishes any nested interval property for it, and the page that built $K$ says why a new field was constructed rather than reusing that one.
The companion failure is FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property, refuted by the same field. Together they are the exact content of the Archimedean hypotheses in clauses 2 and 4 of For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness.

Depends on

Used by

Which of the five completeness properties carry the Archimedean property on their own, and which must be handed it Remark

Dependency tree · two levels

43 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

J. F. Hall, Completeness of Ordered Fields (standard reference, not scraped)
Nested intervals (Wikipedia) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)
Formal power series (Wikipedia) (standard reference, not scraped)