An interesting question has arisen over at this
math.stackexchange
question
about two concepts of *even* in the context of infinite
cardinalities, which are equivalent under the axiom of
choice, but which it seems might separate when choice
fails.

On the one hand, a set $A$ can be even in the sense that it
can be *split into pairs*, meaning that there is a
partition of $A$ into sets of size two, or in other words,
if there is an equivalence relation on $A$, such that every
equivalence class has exactly two elements.

On the other hand, a set $A$ can be even in the sense that
it can be *cut in half*, meaning that $A$ is the union of
two disjoint sets that are equinumerous.

Note that if $A$ can be cut in half, then it can be split into pairs, since if $A=A_0\sqcup A_1$ and $f:A_0\cong A_1$ is a bijection, then $A$ is the union of the family of pairs $\{x,f(x)\}$ for $x\in A_0$. And this argument does not use the axiom of choice.

Conversely, if $A$ can be split into pairs, and if we have the axiom of choice for sets of pairs, then we may select one element from each pair, and $A$ is the union of this choice set and its complement in $A$, which are equinumerous.

Thus, when the axiom of choice for sets of pairs holds, then the two concepts of even are equivalent. Note also that every infinite well-orderable set is even in both senses, and so in ZFC, every infinite set is even in both senses. My question is, how bad can it get when choice fails?

Is there a model of ZF in which every infinite set can be split into pairs, but not every infinite set can be cut in half?

Is there a model of ZF having at least one infinite set that can be split into pairs, but not cut in half?

What is the relationship between the equivalence of the two concepts of even and the axiom of choice for sets of pairs?

distinguishedbijection $A_0 \overset\sim\to A_1$. In the split-into-pairs case, you suppose that your set comes with a partition, and that "every equivalence class has exactly two elements". I've never known how to understand the latter style claim. Certainly you are not saying that "every equivalence class comes with adistinguishedbijection to $\lbrace 0,1\rbrace$, as then I think you have (continued) – Theo Johnson-Freyd Jul 7 '11 at 0:50existenceof a bijection, we can conclude theexistenceof a splitting into pairs. Every bijection can be explicitly converted into a partition into pairs. (If we had a distinguished bijection, that would give a distinguished partition into pairs, but that's not required by the definition of split-into-pairs.) On the other hand, from a partition into pairs and (therefore) the existence of bijections to {0,1} for each of these pairs, I can't conclude much without choice. – Andreas Blass Jul 7 '11 at 2:32