$\text{ZFC}$ proves that each $\aleph\_{n}$ for $n\in \omega$ is a regular cardinal. But it seems without the Axiom of Choice there are many consistent possible choices for cofinality of such cardinals.


**Please introduce some references for any known consistency result with $\text{ZF}$ about possible cofinalities of $\{ \aleph\_n\}\_{n\in \omega}$.** 



Possible Choices for Cofinality of $\aleph_n$ without Choice
Here are some partial answers:


1) By a result of Gitik, all $\aleph\_n$'s can have cofinality $\omega.$


2) the paper "Cofinality and measurability of the first three uncountable cardinals" by Apter, Jackson and Löwe completely solves the problem for $\aleph\_1, \aleph\_2$ and $\aleph\_3.$ The following is taken from their introduction:


"In this paper, we investigate all possible patterns of measurability and cofinality for [$\aleph\_1, \aleph\_2, \aleph\_3$]. Combinatorially, there are exactly 60 such patterns of which 13 are impossible for trivial reasons (e.g., if $\aleph\_1$ is singular, then $\aleph\_2$ cannot have cofinality $\aleph\_1$). In this paper, we prove that the remaining 47 patterns are all consistent relative to large cardinals.'' 


3) The following is appeared in Ioanna Dimitriou's PhD thesis:


**Theorem.** Assume V is a model of ZFC which contains $\omega-$many strongly compact cardinals. For any function $f: \omega \rightarrow 2$ there is a model of ZF in which $\aleph\_{n+1}$ is regular if $f(n)=1$ and singular if $f(n)=0.$


