Is there any set-theoretic axiom that is equivalent to "there exists a well-order on $\mathbb{R}$"?

Question

The question is as the title says: Is there any set-theoretic (ZF) axiom that is equivalent to "there exists a well-order on $\mathbb{R}$"? That is, I want a weaker axiom that axiom of choice that only imposes the statement. Axiom DC does not seem to be the one, so my question.

Answer 1

Nothing particular. Although this sort of question is often hard to answer to the negative, because you can never think about all the options.

As you noted, Dependent Choice is not enough; the same can be said on the Boolean Prime Ideal theorem.

Hardly anything which is not "more or less requiring this explicitly" gives you the wanted result. For example, "There is an infinite set whose power set can be well ordered", or "There exists $\alpha>\omega$ such that $V_\alpha$ can be well ordered" are more or less equivalent to "the reals can be well ordered". But maybe this is what you're looking for.

Answer 2

Here's an equivalent statement which might be of some interest to you:

There is a choice function on the collection of all non-empty sets of reals.

You can prove that this statement is equivalent to the statement that the reals can be well-ordered by following the usual proof that AC is equivalent to the statement that every set can be well-ordered:

1. If you have a well-ordering of the reals, you can define a choice function $f$ mapping $\;\mathscr{P}(\mathbb{R})\setminus\{\emptyset\}\;$ to $\mathbb{R}$ by simply letting $f(x)$ be the least element of $x$ according to the well-ordering.

2. Conversely, if you have such a choice function $f,$ you can get a well-ordering of $\mathbb{R}$ as follows. By Hartogs' theorem, there is a cardinal $\kappa$ so large that there is no 1-1 function from $\kappa$ to $\mathbb{R}.$ Using transfinite induction, define a function $g$ mapping some initial segment of $\kappa$ to $\mathbb{R}$ by setting $g(\alpha)=f(\mathbb{R}\setminus (g''\alpha))$ as long as $\mathbb{R}\setminus (g''\alpha)$ is non-empty. The domain of $g$ is some ordinal $\lambda\le\kappa.$ Also, $g$ is injective on its domain, so in fact we must have $\lambda\lt\kappa.$ Now $\lambda$ is not in the domain of $g,$ so $\mathbb{R}\setminus (g''\lambda)$ is empty. It follows that $g$ is a bijection between $\lambda$ and $\mathbb{R},$ so the natural well-ordering of $\lambda$ induces a well-ordering of $\mathbb{R}.$

[In the argument above, $g''\alpha$ means the range of $g$ on $\alpha,$ or, equivalently, $\{g(\beta)\mid\beta\lt\alpha\}$.]