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Axiom of Choice has many variants like the followings:

  • There is a choice set for every family of non-empty sets.

  • All sets are well-orderable.

Of course in many cases one don't need AC to prove existence of a choice set or well-ordering for a particular family or set. In such cases one can prove these facts in ZF.

Question: What are examples of set theoretic axioms which violate the axiom of choice badly? I mean examples of axioms which imply non-existence of choice sets (or well-orderings) on almost all families of non-empty sets (or sets) which existence of a choice set (or well-ordering) on them is not provable in ZF.

As it stated in comments, $AD$ is certainly a classic example of an anti-choice axiom. However I would like to know about other variants of such axioms and also the intuition behind them which illustrates why these axioms are violating $AC$ badly not merely implying $\neg AC$ which is possible just by producing a single set which has no well-ordering.

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    $\begingroup$ Axiom of Determinacy en.wikipedia.org/wiki/Axiom_of_determinacy $\endgroup$ – Forever Mozart Dec 20 '14 at 1:43
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    $\begingroup$ @ForeverMozart No, that's a bad example. In fact, determinacy implies several nontrivial positive instances of the axiom of choice. $\endgroup$ – Andrés E. Caicedo Dec 20 '14 at 2:28
  • $\begingroup$ @AndresCaicedo (+1) Thanks for illustrating this subtle point. Would you please provide more explanations? $\endgroup$ – user180918 Dec 20 '14 at 3:42
  • $\begingroup$ @AndresCaicedo Can't we interpret the fact that "$AD\vdash \omega_1~\text{is measurable}$" as an evidence of "bad violation" of $AC$ by $AD$? $\endgroup$ – user180918 Dec 20 '14 at 21:03
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Well, you will always have a proper class of families with a choice function, so in what sense do we mean almost all? Here are a few interpretations.

  1. You could say, every family can be extended to one that doesn't have a choice function. But that is really just the failure of $\sf AC$. Fix one counterexample, and use that.

    Other variants like "...by adding a countable family of pairs" equally just mean the failure of the axiom of choice for that type of families.

  2. You could say, the power set of every infinite set cannot be well-ordered. Since $X$ can be well-ordered if and only if there is a choice function from $\mathcal P(X)\setminus\{\varnothing\}$ this means that taking power sets ensures that an infinite set cannot be well-ordered.

    But look closely, this just means that $\Bbb R$ cannot be well-ordered.

  3. You could say, one of the most basic families we know are enumerations of ordinals. So the failure to choose from there is considered quite severe. For example, it could be the case that we cannot choose an enumeration for each countable ordinal uniformly. This is true in the Solovay model, or in models were $\omega_1$ is singular. But similar things are true in models of $\sf AD$ where many successors are singular, while simultaneously

    In fact this can be extended to the Gitik model where each successor cardinal has cofinality $\omega$ (so every limit cardinal too). And in that model we have some sort of absolute failure of countable choice.

  4. You could say, I want to take Gitik's work one step further. One question left open by Gitik is the consistency of Specker's $\sf (K)$ principle stating that $\mathcal P(\kappa)$ is the countable union of sets of size $\kappa$ for every infinite cardinal. In the Gitik model this is false since $\Bbb R$ is not the countable union of countable sets (but rather the countable union of countable unions of countable sets).

    This too, in some (stronger) sense is some sort of absolute failure of countable choice.

  5. You could say, that you want to have countable choice fails in a bizarre way. Douglas Morris constructed a model which has the same cofinality structure as its $L$ (or whichever ground model you started with), but there was no bound on the cardinality of countable unions. Namely, for every ordinal $\alpha$, there is a set $X$ which is the countable union of countable sets, and $\mathcal P(X)$ can be mapped onto $\alpha$.

    Morris did this from an inaccessible, but remarks in the announcement of the result (Notices of AMS 17 (1970), p. 577) that this use can probably be dispensed. This would be a weaker global failure of countable choice, but nonetheless a severe failure. For one, there is no way to extend this model to a model of choice without adding ordinals.

  6. You could say, let's go a different direction altogether. In the absence of choice we define cardinals of non-well orderable sets as those sets of least rank which were realized from that cardinality (namely, $|A|=\{B\mid\text{ there is a bijection }f\colon A\to B\land\operatorname{rank}(B)\text{ is minimal}\}$).

    It is consistent that the axiom of choice fails and you cannot choose representatives for these classes. Namely, there is no function $C$ such that $|C(A)|=|A|$ for all $A$, and whenever $|A|=|B|$ we have $C(A)=C(B)$. It is consistent that such function exists, though. But this is some sort of failure which is hard to quantify in terms of almost all families, but rather "obvious" families.

  7. In a relatively similar way to the previous one, not quite a global failure as a local but strange failure. If you consider Spector's model for having the club filter of $\omega_1$ as an ultrafilter without large cardinals, the axiom of countable choice fails there. The reason is that the countable intersection of clubs of $\omega_1$ is always a club, but the existence of a countably complete ultrafilter implies a measurable.

    The reason is that there is a sequence of sets, each contains a club, but we cannot choose uniformly a club for each one.

The list can probably be extended in several other directions. For my money, I'd go for Gitik's model or theoretically for Specker's $\sf (K)$ (whose consistency, as far as I know, remains open for now).

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  • $\begingroup$ Thanks for this comprehensive answer. As a motivation for this question, I should mention the problem of Kunen inconsistency in the absence of choice. It seems the Axiom of Choice have to be violated badly in $V\models ZF$ to allow producing a non-trivial elementary embedding from $V$ to $V$. Have you ever thought about this problem? What is your idea about such a choice less model? What sort of violation of $AC$ is there in a model of $\exists j:V\rightarrow V$? $\endgroup$ – user180918 Dec 20 '14 at 21:00
  • $\begingroup$ No, I haven't taken this as a serious mathematical investigation. What sort of violation would be in such model? Severe one, of course. Since it means that the universe is in some sense very far from its inner models of $\sf ZFC$. $\endgroup$ – Asaf Karagila Dec 20 '14 at 21:15
  • $\begingroup$ Is there any list of the most famous/important problems of choice-less set theory anywhere? I am curious to know if the problem of Kunen inconsistency without choice is the most important one or not! $\endgroup$ – user180918 Dec 20 '14 at 21:35
  • $\begingroup$ Importance is a matter of opinion. I agree that the Kunen inconsistency problem is an important one to many people, but if we can't prove it for a good period of time? Does it mean that we can start saying that Reinhardt cardinals are consistent with $\sf ZF$? I doubt anyone would say that. We lack sufficient evidence for this sort of statements. There are problems like Ramsey measurability requiring an inaccessible; and there is The Partition Principle; and there are various other problems. All of which can be considered very important. I don't discriminate, I like them all. $\endgroup$ – Asaf Karagila Dec 20 '14 at 21:38
  • $\begingroup$ I am agree. What about list? Is there any list of important choice-less open problems anywhere on the net (or notes)? $\endgroup$ – user180918 Dec 20 '14 at 21:48

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