Splitting infinite sets There are two questions here, an explicit one, and another (more vague) one that motivates it:
I am pretty certain the following should have a negative answer, but at the moment I'm not seeing how to argue about this and cannot locate an appropriate reference. 

In set theory without choice, suppose
  $X$ is an infinite set such that for
  every positive integer $n$, we can
  split $X$ into $n$ (disjoint) infinite
  sets. Does it follow that $X$ can be
  split into infinitely many infinite sets? What would be a reasonably weak additional assumption to ensure the conclusion. 

("Reasonably weak" would ideally be something that by itself does not suffice to give us that $X$ admits such a splitting, but I am flexible.)
This was motivated by a question at Math.SE, namely whether an infinite set can be partitioned into infinitely many infinite sets. This is of course trivial with choice. In fact, all we need to split $X$ is that it can be mapped surjectively onto ${\mathbb N}$. 
However, without choice there may be counterexamples: A set $X$ is amorphous iff any subset of $X$ is either finite or else its complement in $X$ is finite. It is consistent that there are infinite amorphous sets. If $X$ is infinite and a finite union of amorphous sets, then $X$ is a counterexample. The question is a baby step towards trying to understand the nature of other counterexamples. 
Note that any counterexample must be an infinite Dedekind finite (iDf) set $X$. One can show that for any iDf $X$, ${\mathcal P}^2(X)$ is Dedekind infinite. For any $Y$, if ${\mathcal P}(Y)$ is Dedekind infinite, then $Y$ can be mapped onto $\omega$ (this is a result of Kuratowski, it appears in pages 94, 95 of Alfred Tarski, "Sur les ensembles finis", Fundamenta Mathematicae 6 (1924), 45–95). As mentioned above, our counterexample $X$ cannot be mapped onto $\omega$, so ${\mathcal P}(X)$ must also be an iDf set. 
The second, more vague, question asks what additional conditions should a counterexample satisfy.
 A: Define a permutation model of ZFA as follows.  Starting as usual (Ch 4 of Jech's Axiom of Choice) from a well-founded model $\mathcal M$ of ZFAC with infinite set $A$ of atoms, let $G$ be the group of all permutations of $A$; so $G$ can be identified with the group of all automorphisms of $\mathcal M$.  For each finite partition $T$ of $A$, let $G_{(T)}$ be the group of permutations in $G$ which fix each element of $T$ (meaning $\sigma\in G_{(T)}$ iff for each $B\in T$ we have that $b\in B$ implies $\sigma b\in B$).  Let $\mathcal F$ be the set of subgroups of $G$ which contain $G_{(T)}$ for some finite partition $T$; then $\mathcal F$ is a normal filter of subgroups of $G$, and contains the stabilizer subgroup of each atom in $A$.  As usual, a set or atom $x\in\mathcal M$ is called symmetric if its stabilizer subgroup is a member of $\mathcal F$, and we let $\mathcal N$ be the class of hereditarily symmetric elements of $\mathcal M$.
Then $\mathcal N$ is a model of ZFA providing a counterexample.  The model $\mathcal N$ has all the finite partitions of $A$ found in $\mathcal M$, but every infinite partition of $A$ into non-singletons would fail to be symmetric.
