Many people find ACC more intuitive than AC ("Pick something from the first set, then something from the second set, then...) and it also doesn't lead to "controversial consequences" (See for eg: Peculiar examples with Axiom of Countable Choice ?)

My question is:
What are the consequences for Set Theory if we replace AC by ACC -as in ZF and ACC assumed true but cannot assume AC ?

1) Are all ZFC ordinals - $\aleph_1$, $\aleph_\omega$, aleph and beth fixed points etc - still well-defined ?
2) Is the Continuum Hypothesis still undecidable ?
3) Are there any striking changes to Large Cardinal properties ? (For eg: "The smallest measurable cardinal can be equal to the smallest strongly inaccessible cardinal")

PS: If the question is too broad, I'd be very happy to be referred to a book/paper.

Edited: Realized from a comment below that ZF + ACC + Not AC was the system I had in mind, otherwise question 2 becomes trivial.

  • 9
    $\begingroup$ I thought it was automatic that, if you weaken an axiomatic theory, then anything that was undecidable before is still undecidable? $\endgroup$
    – bof
    Nov 10, 2015 at 1:43
  • 3
    $\begingroup$ Following on what bof said in his first comment: en.wikipedia.org/wiki/Hartogs_number $\endgroup$
    – Todd Trimble
    Nov 10, 2015 at 1:46
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    $\begingroup$ Surely, if CH could be proved or disproved with ACC, then it could be proved or disproved in set theory with the full AC? So why is 2) a question? $\endgroup$
    – bof
    Nov 10, 2015 at 2:10
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    $\begingroup$ None of the usual basic theory of alephs uses choice, only the assertion that every cardinal is an aleph. $\endgroup$ Nov 10, 2015 at 2:13
  • 5
    $\begingroup$ 1) Yes. 2) Yes. 3) Yes. How could $\mathsf{CH}$ not be undecidable, by the way, if it is undecidable in the stronger theory $\mathsf{ZFC}$? Any standard reference should address this. For instance, Jech's set theory book. $\endgroup$ Nov 10, 2015 at 2:26

1 Answer 1


First of all, the axiom of countable choice says that given a countable family of non-empty sets, you can choose from each set simultaneously. If you want to choose from one, then from another, then from another, and so on you need a strictly stronger form of choice called Dependent Choice, abbreviated as $\sf DC$.

To your questions, the definition of the $\aleph$ numbers uses absolutely no choice, although the proof that any of them is regular does use the axiom of choice (well, except $\aleph_0$). So it is consistent, for example, that $\aleph_1$ is singular. But assuming $\sf DC$ will prevent that. Whether or not you can have every cardinal $\geq\aleph_2$ singular with $\sf ZF+DC$ is still open.

The definition of $\beth$ numbers, on the other hand, goes out the window. The axiom of choice is equivalent to saying that the power set of a well-ordered set is well-ordered. So if the axiom of choice fails, there will be some ordinal $\alpha$ whose power set cannot be well-ordered. This means that at some point $\beth$ cardinals will not be $\aleph$ numbers. You can still talk about $\beth$ numbers, of course, as iterated powers of $\omega$, but that gives you significantly less information in that sense.

As for the continuum hypothesis, as noted by others it is still unprovable. But now you even get different forms of the continuum hypothesis. $2^{\aleph_0}=\aleph_1$ is no longer equivalent to "Every uncountable set of reals is equipollent with the reals themselves". Indeed, even without $\sf DC$, it is consistent that $2^{\aleph_0}$ and $\aleph_1$ are incomparable, and every uncountable set of reals has cardinality $2^{\aleph_0}$.

Finally, for large cardinals, most properties which are reserved for very large cardinals can be made compatible with $\aleph_1$ and $\sf ZF+DC$. For a more complete survey, look at What sort of large cardinal can $\aleph_1$ be without the axiom of choice?.


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