Two equivalent formulations of the axiom of choice are:

  1. Every family $(X_i)_{i \in I}$ of pairwise disjoint nonempty subsets of a set $X$ has a choice function.
  2. Every family $(X_i)_{i \in I}$ of nonempty subsets of a set $X$ has a choice function.

However, the usual proof of the implication (1) → (2) replaces the set $X$ with the set $X \times I$ and extracts a choice function for $(X_i)_{i \in I}$ from a choice function for the pairwise disjoint family $(X_i\times\lbrace i \rbrace)_{i \in I}$ of nonempty subsets of $X \times I$. Since $X \times I$ can be much larger than $X$, there is no reason to believe that (1) → (2) for a fixed set $X$.

For a given set $X$, (2) has a maximal instance where the family $(X_i)_{i \in I}$ consists of all nonempty subsets of $X$. We therefore see that (2) is equivalent to:

  • There is a choice function $\mathcal{P}(X)\setminus\lbrace\varnothing\rbrace\to X$.
  • The set $X$ is wellorderable.

For a given set $X$, (1) is equivalent to:

  • Every surjection $q:X \to Y$ has a right inverse.
  • Every equivalence relation on $X$ has a transversal.

It appears that (2) is indeed stronger than (1) for a fixed set $X$ and I feel that this should be well known, but I don't recall a model of ZF (or ZFA) where some set $X$ satisfies (1) but not (2). Does anyone know such a model? A model of ZF where $X = \mathbb{R}$ satisfies (1) but not (2) would be most interesting.


Suppose that $X$ is a strongly amorphous set, that is an amorphous set that every partition has only finitely many non-singletons parts. It is clear that there is no choice function from every family of non-empty subsets of $X$, as $X$ cannot be split into two disjoint infinite sets.

If $\cal F$ is a pairwise-disjoint family of subsets of $X$ then without loss of generality $\cal F$ is a partition of $X$ (otherwise simply add the complement of $\bigcup\cal F$). If this partition is finite, then of course there is a choice function, however if the partition is infinite then all but finitely many sets in the partition are singletons, and the choice function is trivial (choose from the non-singletons, and the singletons allow only one choice).

Some remarks on the case when $X=\mathbb R$ (which may not be consistent, though):

  • Since every countable family can be made disjoint this implies countable choice for sets of real numbers.
  • In fact, this means that every ordinal $\kappa$ which $\mathbb R$ can be mapped onto, can be mapped into $\mathbb R$, so there are $\aleph_1$ many real numbers. In particular $\aleph(\mathbb R)=\Theta(\mathbb R)$ (where $\Theta$ is the least ordinal that there is no surjection onto it from $\mathbb R$).
  • Every family of at most $\frak c$ many sets must have a choice function. We can consider a bijection of $\mathbb R$ with $\mathbb R^2$, and if $\lbrace A_r\mid r\in\mathbb R\rbrace$ is a family, then the preimage of $\lbrace r\rbrace\times A_r$ form a disjoint family.
  • To complement the above one, every partition of $\mathbb R$ has size of at most $\frak c$ since there exists a choice function back into $\mathbb R$.

The above rule out the 'usual' models where $\mathbb R$ is not well-orderable, e.g. Feferman-Levy, Solovay, ZF+AD, Cohen's first model. It could be very well that the above already show that there cannot be a model in which $X=\mathbb R$ but I cannot see a reason why (or a model where it is true).

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  • $\begingroup$ Great! Where is the model? I don't think $\mathbb{R}$ have a strongly amorphous subset though, right? $\endgroup$ – François G. Dorais Jul 17 '12 at 21:22
  • $\begingroup$ Any model with a strongly amorphous set would do, Fraenkel's first model, for example. Yes, subsets of $\mathbb R$ cannot be amorphous since amorphous sets cannot be well-ordered. $\endgroup$ – Asaf Karagila Jul 17 '12 at 21:30
  • $\begingroup$ Ah, yes, I see. Those are the only partitions definable in the language of equality. I still want to separate (1) and (2) for $X = \mathbb{R}$. Any thoughts? $\endgroup$ – François G. Dorais Jul 17 '12 at 21:33
  • $\begingroup$ @Francois: I'm not sure. Linearly ordered sets can get quite vicious. The two reasonable models to start with are Cohen's first model with a D-finite set of reals which has a strong-like property (partitions into finite sets are almost all singletons), but this has the problem that there is no choice function from partitions into infinite sets; and the reals as a countable union of countable sets, which has the problem that the countable union can be made disjoint and I am fairly certain we could not choose from this partition. $\endgroup$ – Asaf Karagila Jul 17 '12 at 21:57
  • $\begingroup$ @Francois: I have an idea how to get $X=\mathbb R$, but it utilizes an idea that I need to write in details anyway, if it checks out I'll post an update. $\endgroup$ – Asaf Karagila Jul 18 '12 at 11:52

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