For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness

Let $F$ be an ordered field, with the five properties (LUB), (MCT), (NIP), (BW), (CC) and the Archimedean property (ARCH) as in The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness and Archimedean ordered field. The following five statements about $F$ are equivalent:

1. (LUB);
2. (ARCH) and (NIP);
3. (BW);
4. (ARCH) and (CC);
5. (MCT).

Moreover each of (LUB), (BW) and (MCT) implies (ARCH) on its own, so in statements 1, 3 and 5 the Archimedean property is a consequence rather than a hypothesis.

The Archimedean hypothesis in statements 2 and 4 may not be dropped. It is not an artefact of the proof: the nested interval property without it does not imply (LUB) (FALSE: the nested interval property alone implies the least-upper-bound property), and neither does Cauchy completeness without it (FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property). Both are refuted by the same witness, the formal Laurent series field $\mathbb{R}((t^{-1}))$.

The equivalence is proved as a single cycle $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 1$, each arrow being one lemma of this page.

• What the cycle costs. Seven lemmas suffice for the whole equivalence, because a single cycle through all five statements yields every implication between them, and the arrangement is chosen so that no lemma has to carry an Archimedean hypothesis it cannot discharge. Statement 3 is deliberately the hinge: (BW) is the one property that both implies (ARCH) and is implied by a nested interval argument, so the cycle can enter and leave it without an extra hypothesis.

• Read as a statement about $\mathbb{R}$, the theorem says that the five familiar theorems of a first analysis course are not five theorems but one, and that the least-upper-bound axiom could have been replaced by any of the other four (with (ARCH) alongside, where required). This library takes (LUB) as the axiom (Complete ordered field (least-upper-bound property)) and proves the others from it on earlier pages; nothing here re-proves them for $\mathbb{R}$, and nothing here may be cited as a proof about $\mathbb{R}$ that is not already there.

• The two failures are genuinely different from the three successes. (NIP) and (CC) are both statements about sequences whose data are already close together, and neither of them ever produces a new element far away; that is why an infinitesimal layer can be added to a field without disturbing them, and why the naturals can stay bounded. (LUB), (BW) and (MCT) each quantify over an object that is only assumed bounded, so each of them can be tested against the canonical naturals themselves, and each fails at once when those are bounded. Which of the five completeness properties carry the Archimedean property on their own, and which must be handed it develops this.