Axiom of Choice and Completeness of the Reals

Question

I have a question regarding the Completeness axiom of R. Does It have a relation with the axiom of choice? I mean, does the axiom of choice implies the Completeness axiom? Or are they independent?

Answer 1

Choice has nothing to do with the completeness of $\mathbb{R}$. To be precise, none of the constructions of an Archimedean complete ordered field or proofs of the uniqueness of such up to isomorphism (that I'm aware of, anyways) invoke the axiom of choice in any way.

• Let me clarify the language of the above paragraph a bit, since it may seem odd at first. Since there are multiple reasonable ways to construct $\mathbb{R}$ in the first place (Dedekind cuts, Cauchy sequences, Eudoxus' functions, and many others) the whole notion of "constructing $\mathbb{R}$" in the first place is a bit more slippery than one might expect. In my opinion, the right interpretation of this phrase is: "proving that there is exactly one Archimedean complete ordered field up to isomorphism." We then use the symbol "$\mathbb{R}$," the term "real numbers," and so on to refer to any such field, the point being that uniqueness-up-to-isomorphism makes this ambiguity benign in a precise sense. Meanwhile, notice that completeness is explicitly part of the definition of $\mathbb{R}$ in this approach - so "constructing $\mathbb{R}$" includes "showing that $\mathbb{R}$ is complete" as a necessary step along the way.

Now admittedly a bit of set theory does come into play in constructing the reals - necessarily so, in a precise sense - but it's the powerset axiom, not the axiom of choice. Specifically, the things we use to build $\mathbb{R}$ from $\mathbb{Q}$ are "higher-type" objects - sets of rationals in the construction via Dedekind cuts, sets of sequences of rationals in the construction via Cauchy sequences, and so on - and showing that there is a set of all such objects in the first place requires the powerset axiom.

It's separately worth noting - and this was mentioned by Patrick Stevens in his comment above - that there is a computability theoretic aspect to completeness of $\mathbb{R}$. Specifically, there is a computable increasing bounded sequence of rational numbers whose least upper bound is not a computable real number. Consequently, in a precise (if somewhat weak) sense we have to go "outside the computable universe" to see completeness.

Answer 2

They're unrelated. The completeness of the reals is property $4$ here, stating a nonempty set of reals bounded above has a least upper bound.