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.



Distinguishing two local versions of the axiom of choice
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).




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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).


