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.

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    $\begingroup$ If X is infinite, then 2^X can be mapped onto omega. This is provable in ZF and has nothing to do with X being Dedekind finite. $\endgroup$
    – user5810
    Jan 5, 2011 at 2:18
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    $\begingroup$ @Ricky: Sure. That was poorly phrased. I've changed the sentence into what I really meant. $\endgroup$ Jan 5, 2011 at 2:53
  • $\begingroup$ Could one of the compactness theorems of logic be used here? $\endgroup$ Jan 6, 2011 at 7:47
  • $\begingroup$ @Michael: I am not sure I see how. Given $X$ you would need to devise a theory (presumably with infinitely many relational symbols that would play the role of the partition) for which there is a model with universe equipotent to $X$. This seems very delicate (we do not have the Löwenheim–Skolem theorem without choice). Moreover, we need to ensure that the theory has no models of size $Y$ for any amorphous $Y$. Plus, we would need to anticipate (to set up the language!) the infinite set $Z$ such that $X$ is a union indexed by $Z$ of infinite sets (I think $Z$ doesn't need to be countable). $\endgroup$ Jan 6, 2011 at 7:58
  • $\begingroup$ @Michael: Given $X$, the other possibility I can think of would be to have the language contain a relational symbol $R$ and a function symbol $f$, and set up a theory that would have a model of the form $X\sqcup Y$ where $R$ is interpreted by $X$, $Y$ is infinite, and $f\upharpoonright X$ is a function from $X$ onto $Y$ with the preimage of each element of $Y$ being infinite. Again, the lack of an appropriate version of Löwenheim–Skolem is a serious issue here. $\endgroup$ Jan 6, 2011 at 8:01

1 Answer 1


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.

  • $\begingroup$ Hi Eric. Welcome to MO! Many thanks; I definitely need to become more comfortable with this technique. $\endgroup$ Jan 9, 2011 at 15:14
  • $\begingroup$ Hello and thanks. Yes, if you like this kind of question, then the technique is worthwhile. But permutation groups may not be necessary. If you prefer, you could think about $L(S)$, where, in a model of ZFAC, $S$ is the set of finite partitions of the set of atoms. (Does anyone in fact prefer to think of it that way?) So far, I don't have an answer to your second question. $\endgroup$
    – Eric Hall
    Jan 9, 2011 at 16:26

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