Existence of model of ZF without AC, but with many choice function Question 1: Does there exist models of the Zermelo-Fraenkel set theory without the axiom of choice, but such that every indexed family of non-void sets whose index set has a well-orderable cardinal admits a choice function ?
Question 2: The same as question 1, with "a well-orderable cardinal" replaced by "a linearly ordered cardinal".
Gérard Lang
 A: For the first question:
Yes. It is known that the axiom "Every well-orderable family of non-empty set has a choice function" implies $DC$ but not $DC_{\aleph_1}$.
You can find the proofs in Jech's "The Axiom of Choice" and in Felgner's "Models of ZF-Set Theory".
I am not sure about the second question, but I believe the answer should be "yes" as well.
In Consequences of the Axiom of Choice you can find the following principles:


*

*Form 1 is the axiom of choice.

*Form 202 is the existence of a choice function for linearly ordered families of non-empty sets.

*Form 40 is the existence of a choice function for well-ordered families of non-empty sets.


Entering those three numbers in the table show that $1\implies 202\implies 40$ and neither implications is reversible. Although in the case of $202\implies 1$ this is in a weaker axioms system without regularity.

Edit: I had a bit more time now, so I went to chase after the cited source for Form 202:

Truss, J. The axiom of choice for linearly ordered families. Fund. Math. *99 (1978), no. 2, 133–139. (MR480029)

Here is an excerpt from the review:

The axiom of choice for linearly ordered families asserts that any linearly ordered family of nonempty sets has a choice function. The author shows that in ZF this statement implies the full axiom of choice, while in FM (ZF without the axiom of foundation) it does not.

So it seems that in ZF, requiring for linearly ordered families is enough to require the full axiom of choice.
