How well-behaved can well-orders on $\mathbb{R}$ be?

Question

Well-orders on $\mathbb{R}$ are interesting because they witness the "alephness" of the cardinality of the continuum, and they're therefore connected to the continuum problem. It seems reasonable, therefore, to ask: "How well-behaved can well-orders on $\mathbb{R}$ be?"

This is too vague for the website, so let me instead ask:

Question. Does ZFC prove the following?

There exists a well-order $\unlhd$ on $\mathbb{R}$ such that for all $x,y \in \mathbb{R}$ with $x < y$, there exists an ("intermediate") real number $i \in \mathbb{R}$ satisfying:

$$x < i < y, \qquad x \unlhd i, \,y \unlhd i$$

If not, does ZFC refute this claim?

Answer 1

Yes, such a well-order exists (in ZF) as long as there exists any well-ordering of $\mathbb{R}$. Indeed, let $\kappa$ be the least ordinal which is in bijection with $\mathbb{R}$ (also known as the cardinal of $\mathbb{R}$), let $f:\mathbb{R}\to\kappa$ be a bijection, and define $x\unlhd y$ to mean $f(x)\leq f(y)$. By minimality of $\kappa$, for any $\alpha\in\kappa$, there are fewer than $|\mathbb{R}|$ ordinals less than $\alpha$. So for any $x,y\in\mathbb{R}$ with $x<y$, there are fewer than $|\mathbb{R}|$ elements $z\in\mathbb{R}$ such that $f(z)\leq f(y)$ or $f(z)\leq f(x)$. In particular, since $|(x,y)|=|\mathbb{R}|$, there must be an $i\in (x,y)$ such that $f(i)>f(x)$ and $f(i)>f(y)$, i.e. such that $x\unlhd i$ and $y\unlhd i$.

I would remark that a huge part of the utility of the theory of well-orderings comes from the fact that if you can well-order a set, you can well-order it such that each initial segment of the well-ordering has strictly smaller cardinality, as we did with $\mathbb{R}$ here. This allows you to do all sorts of inductive arguments where you use the fact that at each intermediate step of your induction, you have only handled a "small" subset of your set so far.

Answer 2

Here's another proof. Let $\prec_0$ be a well-ordering of $\mathbb{R}$ and $\prec_1$ be a well-ordering of $\mathbb{Q}$ of order-type $\omega$ (these are constructed in proofs that $\mathbb{Q}$ is countable), and define $\prec$ as follows: $r\prec s$ iff

• $r\not\in\mathbb{Q}$ but $s\in\mathbb{Q}$, or

• $r, s\not\in \mathbb{Q}$ and $r\prec_0s$, or

• $r, s\in\mathbb{Q}$ and $r\prec_1s$.

Basically, we take $\prec_0$, move the rationals to the front, and order them as $\prec_1$.

Then fix $x<y$. There are infinitely many rationals in between them; since any rational is $\succ_1$ at most finitely many rationals, one of these must be $\succ_1$ both $x$ and $y$.