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This question is prompted by a paper by Andre Kornell that just appeared on the arXiv. A large portion of the paper is devoted to showing that a surprising amount of ordinary mathematics can be developed in Chang's model. Although the axiom of choice does not hold in this model, weaker choice principles are available which apparently suffice for most mainstream uses.

The development is motivated by the author's view that a general theory of "quantum sets" works better in Chang's model. I suppose the fact that every set of reals is Lebesgue measurable could already be cited as a way in which the model is better from the point of view of mainstream math.

Has there been any previous work in this direction, specifically about ordinary mathematics in Chang's model? (Comments about what is known to follow from known properties of the model, e.g., the axiom of determinacy plus dependent choice, are also welcome.)

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    $\begingroup$ Interesting. Is there any reason why Kornell uses Chang's model $L(\mathrm{Ord}^\omega)$ instead of the "more popular" $L(\mathbb{R})$? At some point he states that his results work in $ZF + AC_{ae} + DC + LM + BP + PSP$, which is much weaker than $L(\mathrm{Ord}^\omega) \models AD^+ + DC + AC_{\mathbb{R}}$. That leads me to think that Kornell might be happy with $L(\mathbb{R})$. (By the way, the axiom of choice can hold in Chang's model but it doesn't have to, and the axiom of choice is excluded in Chang's model by some large cardinal hypotheses.) $\endgroup$
    – François G. Dorais
    Feb 6, 2015 at 16:16
  • $\begingroup$ @François: I can't speak for Andre, but I would also be interested in answers to this question for $L({\mathbb R})$. Andre assumes the existence of a proper class of Woodin cardinals that are limits of Woodin cardinals (!) to nail down the properties of $L({\rm Ord}^\omega)$. $\endgroup$
    – Nik Weaver
    Feb 6, 2015 at 16:47
  • $\begingroup$ I'm not sure what "ordinary mathematics" might mean. Lots of independent papers can be cited that investigated things like $\sf DC$ (it is equivalent to the Baire Category theorem) and $\sf LM$ or $\sf BP$ which imply automatic continuity (at least in the presence of $\sf DC$), and so on and so forth. $\endgroup$
    – Asaf Karagila ♦
    Feb 6, 2015 at 17:23
  • $\begingroup$ @Nik, for the analogous result for $L(\mathbb{R})$ you don't need quite so much (just one limit of Woodins, if I recall correctly). There is much more literature on $L(\mathbb{R})$ than there is on $L(\mathrm{Ord}^\omega)$. This is true at least in the literal sense, though one would expect many of the $L(\mathbb{R})$ results to lift to $L(\mathrm{Ord}^\omega)$ after some adjustments... $\endgroup$
    – François G. Dorais
    Feb 7, 2015 at 0:47
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    $\begingroup$ Seems I am overusing MO - when visiting the link to that paper in the arXiv I felt strong urge to upvote :D $\endgroup$
    – მამუკა ჯიბლაძე
    Feb 7, 2015 at 7:58

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I don't have enough reputation to comment, so I'm posting my reply to François as an answer. It is possible that these results can be established from $ZF+AC_{ae}+DC+LM+BP+PSP$, but I don't make this claim. I work in a transitive model of this theory that is closed under countable sequences, or equivalently under countable unions, which implies that certain functional-analytic properties, such as the completeness of a metric space, are absolute. We can often verify a familiar fact by appealing to absoluteness, and this is typically much faster than checking its proof down to first principles. For most individual results, this is a matter of convenience, but for research in an established research area, this is effectively a necessity.

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  • $\begingroup$ Welcome to MO! Do you ever need countable sequences built outside the model? If so, it would be interesting to know where (even if the use is not strictly necessary, in fact it could be even more interesting if it's not actually necessary). $\endgroup$
    – François G. Dorais
    Feb 7, 2015 at 7:10
  • $\begingroup$ What do you mean when you say "built outside the model"? $\endgroup$
    – Andre Kornell
    Feb 7, 2015 at 7:37
  • $\begingroup$ Every model of ZF is closed under internally constructed countable sequences. I'm wondering why you need external countable sequences for (especially since you're assuming DC). $\endgroup$
    – François G. Dorais
    Feb 7, 2015 at 7:39
  • $\begingroup$ Here's a simple example. Every bounded functional on a Hilbert space is of the form $\eta \mapsto \langle \xi| \eta \rangle$ for some $\xi$. I can verify this fact in the Chang model without looking at its proof, by appealing to absoluteness. $\endgroup$
    – Andre Kornell
    Feb 7, 2015 at 7:46
  • $\begingroup$ (I recall that the usual proof of this fact appeals only to $ZF+DC$.) $\endgroup$
    – Andre Kornell
    Feb 7, 2015 at 7:50

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