False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property

Statement

False claim: every ordered field in which every Cauchy sequence converges, that is, every ordered field with (CC) as in 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 4 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. The witness is again the formal Laurent series field $K = R ((t^{- 1}))$ : every Cauchy sequence in $K$ converges in $K$ , and $K$ has no least upper bound for the set of its own canonical naturals.

This is the sharpest of the failures on this page, because "complete" is the word most often used loosely for both properties at once. In $R$ they coincide; in an ordered field they do not, and the difference is exactly the Archimedean property.

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 sequence in $K$ that is Cauchy in $K$ converges in $K$ (Every Cauchy sequence in $R ((t^{- 1}))$ converges: $K$ is sequentially Cauchy complete); Cauchyness and convergence in an ordered field are as in Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field, and that is exactly (CC) (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 (CC) 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 4 implies clause 1).

Refutation

technique · direct

1.1

$K$ is an ordered field.

1.2

$K$ has (CC): every Cauchy sequence in $K$ converges in $K$ .

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 (CC) 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

Where the thresholds are read is what makes this possible. Cauchyness in $K$ is tested against every positive element of $K$ , including the infinitesimals (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field), so the condition is much stronger in $K$ than the same words read with rational thresholds. It is strong enough that only sequences whose coefficients freeze can satisfy it, and those all converge. Meanwhile the canonical naturals, which are what (LUB) fails on, are not Cauchy at all, so (CC) never gets a chance to see them.
The three properties $K$ has and the three it lacks. It has (CC) and (NIP) in the shrinking form ( $R ((t^{- 1}))$ has the nested interval property for lengths tending to $0$ ) and it is an ordered field; it lacks (LUB), and hence also (BW) and (MCT), each of which would force it to be Archimedean (Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis, The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis).
A reader who wants a single sentence: Cauchy completeness says the field has no holes that a sequence can point at; the least-upper-bound property says it has no holes at all. In a non-Archimedean field a sequence indexed by $N$ is too short to point at the holes.

Depends on

Used by

Dependency tree · two levels

42 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)
Complete metric space (Wikipedia) (standard reference, not scraped)
Completeness of the real numbers (Wikipedia) (standard reference, not scraped)
Formal power series (Wikipedia) (standard reference, not scraped)