Let $\kappa$ be a regular cardinal. If I understand correctly, the proof that the intersection of $<\kappa$ many club subsets of $\kappa$ is a club does not require AC. However, the proof that the club filter on $\kappa$ is $\kappa$-complete does ostensibly require AC because given a sequence of sets in the club filter, we need to pick out a sequence of clubs that they contain in order to show that their intersection is in the club filter. So I'd like to know, in the absence of AC, how strong is the assertion that the club filter on $\kappa$ is $\kappa$-complete? Does it imply $AC\_{\kappa}$, or is it weaker? If weaker, then how much weaker?


Is there any well-known model of ZF without AC in which there is a proper class or better of regular cardinals $\kappa$ such that the club filter on $\kappa$ is $\kappa$-complete?



Completeness of the club filter without AC
If $M$ is an inner model of a forcing extension $V[G]$ of a model $V$ of Choice via a partial order of cardinality $\kappa$ in $V$, then for every regular cardinal $\lambda$ of $V$ greater than $\kappa$, every club subset of $\lambda$ in $V[G]$ will contain one from $V$. So, in $M$, the club filter on $\lambda$ will be $\lambda$-complete for any such $\lambda$, as $V[G]$ can carry out the usual argument with the sets from $M$, and then shrink the result to a set in $V$. A similar situation holds for models of ZF which have set-forcing extensions satisfying Choice. There are many such models of each type in which AC fails, for instance the Feferman-Levy model, in which $\omega\_{1}$ is singular. 


Andres's answer is more interesting, of course. 


