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The regularity of $\omega_{1}$ is one of the most well known facts of set theory. But it seems that in order to prove this simple fact we need the "full power" of mathematics! For example by an interesting result of Gitman, Hamkins, Johnstone if we delete the axiom of power set from $ZFC$ then it is possible that $\omega_{1}$ be a singular cardinal. Now a natural question is:

Question: For which one of the axioms of $ZFC$ like $A$ we have:

$Con(ZFC)\Longrightarrow Con(ZFC-A+cf(\omega_{1})<\omega_{1})$

Particularly is the following statement true?

$Con(ZFC)\Longrightarrow Con(ZFC-Rep+cf(\omega_{1})<\omega_{1})$

Note that in some sense the axioms of power set and replacement are the most "powerful" axioms of $ZFC$. Also they have a "dual" role in mathematics.

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up vote 6 down vote accepted

There is an ambiguity in this kind of question, where you refer to ZFC-A for an axiom A, since what the theory is that you refer to depends on the particular way that you axiomatize ZFC. For example, different axiomatizations that are equivalent in the presence of axiom A might no longer be equivalent when you omit A, and so it could be unclear what is meant by ZFC-A.

This kind of issue was the main point of my paper with Gitman and Johnstone, to which you refer: What is the theory ZFC without power set? In this paper, we prove that a variety of surprising statements are consistent with ZFC-power, if one axiomatizes this theory via the replacement axiom, but a better theory is obtained if one axiomatizes the theory via the collection axiom, plus separation (the point being that this is not equivalent to replacement when one lacks the power set axiom). Our summary conclusion is that one should really use collection+separation rather than replacement, when omitting power set, since otherwise so much goes wrong. Thus, when properly axiomatized, I wouldn't agree that $\omega_1$ can be singular in ZFC - powerset. That situation is only consistent with an improper formulation of the theory.

Regarding your specific question at the end, one doesn't need the replacement axiom to prove that $\omega_1$ is regular; this can be proved in the Zermelo theory plus choice, assuming that $\omega_1$ exists. First, one forms the set of relations on $\omega$ that are well-orders. These can be grouped into equivalence classes under the order-isomorphism relation. Now, by AC, one can pick a representative from each class. If $\omega_1$ is singular, then one gets a countable sequence of such representatives of order type unbounded in $\omega_1$. And one can now glue these orders together to make a single countable order type at least as large as $\omega_1$, a contradiction.

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An exact answer. Thanks for notification about the paper authors. – user36136 Oct 17 '13 at 23:35

Without the axiom of choice it is consistent that $\operatorname{cf}(\omega_1)=\omega$. This was shown by Feferman and Levy back in 1964, right after Cohen published his original results about forcing.

The construction is not difficult, you just collapse the $\aleph_n$'s and take a symmetric models defined by bounded permutations. In the resulting model you have that the real numbers are a countable union of countable sets, which implies that $\omega_1$ is also a countable union of countable ordinals.

Interestingly enough, if you instead collapse all the ordinals, rather than just the cardinals, you get the same result about $\omega_1$, but not about the real numbers. This fact is due to Truss (somewhere in the 1970's).

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Dear Asaf, Thanks for your useful answer. Also I have a personal question: What is your most favorite fact (theorem, phenomena, project) in set theory without $AC$? – user36136 Oct 17 '13 at 22:54
Haha, there are just too damn many of them. :-) – Asaf Karagila Oct 17 '13 at 22:55
Dear friend, say something that can "force" me to enter your realm! :-) – user36136 Oct 17 '13 at 22:58
Well, it is consistent to have the failure of $\sf AC$, but with $\sf DC_\kappa$, for an arbitrary $\kappa$, while having descending chains of cardinals of every ordinal length; and a proper class of incomparable cardinals while at it. I also like the fact that the mixing lemma (from forcing) is equivalent to the axiom of choice. – Asaf Karagila Oct 17 '13 at 23:01
Really interesting! Thanks. – user36136 Oct 17 '13 at 23:12

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