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

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 $K = \mathbb{R}((t^{-1}))$: 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 $\mathbb{R}$ they coincide; in an ordered field they do not, and the difference is exactly the Archimedean property.

• 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 ($\mathbb{R}((t^{-1}))$ 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 $\mathbb{N}$ is too short to point at the holes.