Is there an effective (constructive) well order on reals ?

I know several questions were already asked on this topic, and the answers were very good to this well known problem. My question is more specific :

Is there some axiom $A$ ($A$ is equivalent to "there exists some large cardinal with such properties...") such that one can effectively build a well order on reals in ZFC+$A$ ?

I think there is such $A$ but I did not find any articles or relevant links about it. So if someone can give me a link OR assure me I'm wrong about it and nothing was published about that in the last years OR assure me I'm totally wrong and nothing like that is possible, and I'm a guy who must stop maths asap.

Thanks !

  • $\begingroup$ I think this answer is 'No', since ZFC+'there is no definable definable well-ordering of the reals' is consistent if ZFC is consistent. $\endgroup$
    – Hanul Jeon
    Commented Jan 18, 2013 at 14:05
  • $\begingroup$ Yeah, I deleted my comment. But saying there is a consistent axiom that says "there is no definable well ordering" is not enough (no ?) to say there is no consistent axiom that says "there is one way to define..." $\endgroup$
    – Xoff
    Commented Jan 18, 2013 at 14:11
  • 2
    $\begingroup$ mathoverflow.net/questions/71337/… $\endgroup$ Commented Jan 18, 2013 at 14:14
  • $\begingroup$ No, but it does mean that, in ZFC, there is no definable function which is provably a well-ordering of the reals. @Xoff $\endgroup$ Commented Jan 18, 2013 at 14:23
  • $\begingroup$ Thanks for the math overflow link, this is the answer I needed, and there are links to nice articles. Thank you ! $\endgroup$
    – Xoff
    Commented Jan 18, 2013 at 14:29

1 Answer 1


I'm glad you found the links to the MO question useful. Let me add a couple of complementary remarks (that I should probably add to the answer over there at some point):


Recently, Woodin has been studying an axiom he calls "Ultimate $L$", see for example this MO question, and the links I provide there. The intention of the axiom is to provide a framework that should be consistent with all known large cardinals, and implies that the universe admits a fine structure akin to that of $L$ or, more precisely, to that of the core model.

It is a consequence of the axiom (appropriately formulated) that $\mathsf{GCH}$ holds (in a strong sense, we even have definability). In fact, one can provide an upper bound for the complexity of a well-ordering of the reals, using a predicate for the universally Baire sets. The presence of this predicate can be explained, loosely speaking, as follows: The axiom postulates that the universe is approximated by "mice", and a well-ordering of reals can be traced back to the existence of strategies for successfully comparing these mice, but these strategies can be coded via universally Baire sets.

One is left with the question of whether this is a "reasonable" axiom to add to $\mathsf{ZFC}$. In fact, one can think of it as a "completion." The point is that we should be able to show the consistency of any "natural" theory by showing it holds in some appropriate inner model of a forcing extension of a rank initial segment of $V$. This is expected to be a consequence of an appropriate version of the so-called $\Omega$-conjecture, see for example this paper by Bagaria-Castells-Larson for an introduction. The philosophy advanced here is that all we should care about is the interpretability power of a (set) theory. Two theories that are bi-interpretable are considered just as good, and preferring one over another is a matter of aesthetics more than anything else.


Curiously, I didn't mention in the forcing-axioms part of my answer to the question first linked above that there is some expectation that Martin's maximum, $\mathsf{MM}$, or at least a natural strengthening, should provide us with a (lightface) definable well-ordering of the reals(!).

There's been a steady amount of work towards settling this question, by several first rate set theorists, but we are not there yet.

  • $\begingroup$ I'm not a first rate set theorist, nor even second or third (omegath perhaps). But thank you for your answer, it is very useful to me :) $\endgroup$
    – Xoff
    Commented Jan 18, 2013 at 20:05

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