12
$\begingroup$

This question is partly motivated by Timothy Chow's recent question on the division paradox.

Say that a set $X$ admits a paradoxical partition if and only if there is an equivalence relation $\sim$ on $X$ such that $|X|<|X/{\sim}|$, that is, there is a partition $$ X=\bigcup_{i\in I}X_i $$ of $X$ into disjoint, non-empty sets $X_i$, $i\in I$, for some set $I$ with $|X|<|I|$ (we split $X$ into more pieces than elements). No such partitions are possible if $X$ is finite or, more generally, well-orderable but, in the absence of choice, examples can be found. Famously (as mentioned in Chow's question), in Solovay's model we have $|\mathbb R|<|\mathbb R/E_0|$, where $E_0$ is Vitali's equivalence relation that identifies two reals precisely when their difference is rational. The same example holds under determinacy.

Assume determinacy and work in $L(\mathbb R)$. Suppose that $X$ is not well-orderable (this means that $\mathbb R$ injects into $X$). Does $X$ admit a paradoxical partition?

In comments, Gabe Goldberg suggested that this may be the case, which I also suspect, but I do not see how to proceed. In particular, I do not see how to take advantage of a paradoxical partition of $\mathbb R$ to obtain one of $\mathcal P(\mathbb R)$. (Note that $2^{|\mathbb R|}=2^{|\mathbb R/E_0|}$.)

Working in $L(\mathbb R)$ under determinacy is more or less just a test case, and any suggestions or partial answers under other assumptions are definitely welcome.

Improve this question
$\endgroup$
2
  • $\begingroup$ I'd hazard a guess that by counting maximal antichains (in an appropriate complete Boolean algebra), one might be able to prove that if AC can be forced, then any sufficiently large set does not have a paradoxical partition, or at the very least for no large enough set with "nice properties", which will always be true for large enough $V_\alpha$, this is the case. So, in particular, this will be the case for $L(\Bbb R)$. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 22, 2022 at 17:42
  • 1
    $\begingroup$ Well, that above comment is at least a bit wrong, that the general case can't work. Indeed, Monro showed that in the Cohen model every non well-orderable set can be split into two smaller sets. In this case, if $X^2=X=A+B$, then there is a map from $A+B$ onto $X^2$ so one of $A$ and $B$ maps onto $X$, so they can be paradoxically partitioned. This does not rule out the possibility that no $V_\alpha$ can be paradoxically partitioned, of course, but the above isn't so simple. (Leaving the previous comment for posterity.) $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 22, 2022 at 17:54

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .