Bolzano-Weierstrass implies Cauchy completeness in any ordered field

Let $F$ be an ordered field with the Bolzano-Weierstrass property (BW) of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness. Then $F$ has Cauchy completeness (CC): every Cauchy sequence in $F$ converges in $F$.

No Archimedean hypothesis is needed here, and none is hidden: (BW) already carries the Archimedean property on its own (Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis), but that fact is not used below.

• This is the only implication on the page that is pure bookkeeping, and that is precisely because both of its ingredients were extracted into Sequence basics in an arbitrary ordered field: limits are unique, limits preserve non-strict inequalities, convergent sequences are Cauchy, Cauchy sequences are bounded, and a Cauchy sequence with a convergent subsequence converges and proved there for an arbitrary ordered field. Written out inline it would repeat the boundedness induction and the three-term triangle estimate of that lemma.

• The converse fails: $\mathbb{R}((t^{-1}))$ has (CC) (Every Cauchy sequence in $\mathbb{R}((t^{-1}))$ converges: $K$ is sequentially Cauchy complete) and, being non-Archimedean ($\mathbb{R}((t^{-1}))$ is non-Archimedean, and the monomials $t^{-k}$ are cofinal below its positive elements), fails (BW) by Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis.