Is it possible to explicitly construct a total order in $\mathbb R^{\mathbb R}$?

Question

Is it possible to explicitly construct a total order in $\mathbb R^{\mathbb R}$?

There is a total order in $\mathbb R^{\mathbb R}$ according to Well-ordering theorem. But I'm curious if there's an explicit way.

Answer 1

The existence of a linear ordering on $\mathbb R^{\mathbb R}$ cannot be proven without using some weak form of the axiom of choice.

Cohen constructed a model of ZF (called "the second Cohen model" in Jech's AC book and some other references) where there is a countable set of pairs of subsets of $\mathbb R$ with no choice function. If $P(\mathbb R)$ could be linearly ordered, there would certainly be a choice function on the pairs (just choose the least of the two), so it follows that $P(\mathbb R)$ cannot be linearly ordered in this model. And so neither can $\mathbb R^{\mathbb R}$ because there is (in ZF) a bijection between $P(\mathbb R)$ and $\mathbb R^{\mathbb R}.$

As with many things that don't necessarily exist in ZF, there is a corresponding result that no definable such thing necessarily exists in ZFC. To be more, but still not fully, precise, it is consistent with ZFC that there is no formula in the language of set theory that defines a linear ordering on $\mathbb R^{\mathbb R}.$

It is also consistent with ZFC that there is a definable linear ordering for the reason that if $V=L$ then there is a definable anything (that exists), since there is a definable well-ordering of all the sets and we can define it as the least one. (And although this might not be what anyone's actually looking for, it's "explicit" in the sense that the definition is not very complicated.)

So, the short answer is, "pretty much no, but it's complicated, and what do you mean by 'explicit', anyway?" This is all very similar to the situation with a well-ordering of $\mathbb R,$ about which the same question is more often asked, and you can probably find a bunch of answers here and on math overflow covering more facets in more detail than I've given here, though not everything you read can be guaranteed to transfer gracefully between the two problems, of course. The first one that came up for me was Carl Mummert's answer here.