5
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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:

  1. Form 1 is the axiom of choice.
  2. Form 202 is the existence of a choice function for linearly ordered families of non-empty sets.
  3. 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.

$\endgroup$
  • $\begingroup$ Just out of curiosity, assuming $ZF+AD$ is consistent, is $ZF+AD$ consistent with "there exists a well-orderable uncountable subset of $[0,1]$? $\endgroup$ – Tomek Kania Jan 4 '13 at 16:01
  • 2
    $\begingroup$ Tomek, no. Under ZF+AD the perfect set property is true, and every set of reals is countable or of size continuum. It follows that if the continuum can be well-ordered the perfect set property fails, and so the only well-ordered sets of real numbers are countable in models of ZF+AD. $\endgroup$ – Asaf Karagila Jan 4 '13 at 16:04
  • $\begingroup$ THANK you very much Asaf. The link with Consequences of the axiom of choice is most interesting. Concerning Jech's book, I only found exercises 4.14 and 5.22 relatives to the proof that AC for well-ordered sets does not prove AC; but this is the case where every non-void set of the indexed family is well-orderable, not the case where the index set is itself well-orderable. Gérard Lang $\endgroup$ – Gérard Lang Jan 4 '13 at 17:02
  • $\begingroup$ Dear Gerhard, check Chapter 8 in Jech's book. $\endgroup$ – Asaf Karagila Jan 4 '13 at 18:08
  • $\begingroup$ Dear Asaf, thank you very much for going until Truss article. Alas, Fund Math seems to be no more freely accessible. Anyway, this gives a most interesting result, notably considering that (as results of Jech's formidable book) the Ordering Principle does not imply the Prime Ideal Theorem, and so also not the Axiom of Choice. Moreover,following your indications, I found theorem 8.12 about failing of the Dependent Choice and nevertheless obtaining AC(l) and W(l) for any cardinal l<k, where k is a regular cardinal. Gérard Lang $\endgroup$ – Gérard Lang Jan 6 '13 at 9:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.