In $\sf ZF$, we have that the axiom of choice is equivalent to:

For all sets $X$, and for all proper classes $Y$, $X$ inject into $Y$


For all sets $X$, and for all proper classes $Y$, $Y$ surject onto $X$

To see that those are indeed equivalent to choice we have for one direction to inject a set $X$ into $Ord$ and this will give well ordering for $X$(and because $Ord$ well ordered, we can easily construct an injective from $X$ to $Ord$ using a surjective from $Ord$ to $X$)

To see that the other direction is true, take a set $α$ and a class $Y$, because we are assuming $\sf AC$ we may assume WLOG that $α∈Ord$. Then we may use induction to create a sequence $(x_β)$ of ordinals such that for $β<γ$ we have $Y∩V_{x_β}\subsetneq Y∩V_{x_γ}$, then we look at $V_{x_α}$, and by well ordering it find an injective $α→Y$(and surjective $Y→α$).

In the proof use relied heavily on the axiom of foundation, so we can ask are those 3 equivalent in $\sf ZF-\mbox{regularity}$?

When talking with @Wojowu he told me that his intuition told him that $\sf AC$ is not equivalent to the other 2, saying that he thinks that there is a model of $\sf ZFC-\mbox{regularity}+\mbox{a proper class of atoms}+\mbox{only finite sets of atoms}$, in which case no infinite set inject into the class of atoms, but after searching I couldn't find any reference to such model. My questions:

If such model exists, can someone direct me to a reference, or explain it's construction? If not, how those 2 behave in $\sf ZF-\mbox{regularity}$?

What about the other 2? Does the surjective version implies the injective version in $\sf ZF-\mbox{regularity}$?

  • 1
    $\begingroup$ math.stackexchange.com/questions/1337583/… might be helpful? $\endgroup$
    – Asaf Karagila
    Apr 25, 2019 at 22:19
  • $\begingroup$ The notation $\sf ZF^-$ usually means "without power set". I don't think there is a very standard notation for "without regularity", but $\sf ZF_0$ was somewhat prevalent in "the old days". $\endgroup$
    – Asaf Karagila
    Apr 26, 2019 at 15:38
  • $\begingroup$ @AsafKaragila in "kenneth kunen set theory an introduction to independence proofs" he uses $\sf ZF^-$ to $\sf ZF$-regularity, I will change it to "$\sf ZF$-regularity" in the question to make it clearer. $\endgroup$
    – Holo
    Apr 26, 2019 at 16:03
  • $\begingroup$ Is that the new one or the old one? $\endgroup$
    – Asaf Karagila
    Apr 26, 2019 at 16:04
  • $\begingroup$ @AsafKaragila New edition $\endgroup$
    – Holo
    Apr 26, 2019 at 16:07

1 Answer 1


The results appear in Jech's "The Axiom of Choice" in the problem section of Chapter 9 (Problems 2,3, and 4).

Indeed, it is easy to see that the injections into classes imply the surjections from classes which imply choice. Exactly by means of the class of ordinals. So the point is to separate the others.

And if we have a proper class of atoms whose subsets are all finite, then it is a class which does not map onto $\omega$, but every set has only finitely many in its transitive closure, so it can be well-ordered.

The last model is described well in Jech, this is Problem 4 in the aforementioned reference, and the key point is that the atoms are indexed by countable sequences of ordinals, so that there are always surjections onto every set, but there is no $\omega$ sequence of atoms, which form a proper class, so there is no injection from any infinite set into the class of atoms. (And indeed, that implies all sets of atoms are finite.)

  • $\begingroup$ Hi Asaf, thanks for answering, you wrote "the atoms are indexed by countable sequences of ordinals, so that there are always surjections onto every set but there is no omega sequence", how does the first and second part works together? And how does it gives you surjective onto a set like omega2? $\endgroup$
    – Holo
    Apr 26, 2019 at 9:59
  • $\begingroup$ Well, map the atom indexed by the constant sequence $\alpha$ to $\alpha$. Of course, you need to show that this map is stable under permutations, which it is, if you extend it "correctly" (i.e. close this class function under permutations to determine its actual domain). $\endgroup$
    – Asaf Karagila
    Apr 26, 2019 at 10:01
  • $\begingroup$ Oh, each atom as an omega sequence of ordinals, I understand now, thanks! $\endgroup$
    – Holo
    Apr 26, 2019 at 10:32

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