Can we have an infinite sequence of paradoxical partitioning below power set?

Question

Can we have the following in absence of choice?

$\exists f [f= \{P_i: i \in \omega\} \land \forall i \in \omega ( P_{i+1}|P_i \land |P_{i+1}| > |P_i| \land |P_i| < |\mathcal P(P_0)|)]$

where $``|"$ denotes "is a partition of".

Answer 1

Yes. You can have an arbitrarily long sequence of paradoxical partitions. Pick your favorite ordinal $\alpha>1$, in your case $\omega$.

Start with a model of $\sf CH$, for simplicity purposes, then add $\omega_\alpha\times\omega$ Cohen reals, and permute each $\omega$ block so that it is Dedekind-finite, but do not permute the sequence of blocks. This gives you a surjection from the reals onto $\omega_\alpha$, but no injection from $\omega_2$ into the reals.

Now partition the reals into $2^{\aleph_0}+\aleph_2$ parts, and then partition that one (by partitioning the part of size continuum) into $2^{\aleph_0}+\aleph_3$ parts, and so on. Since all of these partitions are derived from a fixed surjection from the reals onto $\omega_\alpha$, the sequence of these partitions is also in the model.