10
$\begingroup$

I'm specifically assuming that we have replacement instead of collection; collection breaks things (because then there is a set that contains a map from $n$ to $C$ for every $n\in\mathbb N$, and you can look within that set to get an injection from an infinite subset of $C$ to some infinite set, so $\aleph(C)$ couldn't have been $\aleph_0$.

The construction that seems to work in ZFCA is to take a model of ZFA with infinitely many atoms and then take the direct limit of $L(A_n)$, where $A_n$ is the set of the first $n$ atoms. This seems to be a model of ZFCA, because every set belongs to $L(A_n)$ for some $n$ and therefore inherits a well-ordering from that (as $L(A_n)$ still has a global choice function definable from parameters).

I can't see how replacement fails where collection did, because I can't think any $\varphi$ such that $\forall n\exists!x\varphi(n,x)$ and $\varphi(n,x)$ implies that $x$ is an injection from $n$ into the class of atoms, or anything else that would cause the union of all such $x$s to not fit into any $L(A_n)$.

This of course cannot be extended to a model of GBCA, because the existence of a proper class that is smaller than the universe prevents the existence of a global choice function.

If the construction works and can be extended to ZFC, it does prove that the Hartogs number of classes is interesting in NBG even with the axiom of choice (NBG being conservative over ZFC).

$\endgroup$
7
  • $\begingroup$ Are you talking about a version of ZFCA where the collection of atoms do not need to form a set? $\endgroup$ May 3, 2023 at 23:25
  • 1
    $\begingroup$ Conservativity of GBC over ZFC specifically refers to statements about sets -- GBC can prove things about proper classes that ZFC can't, so your last sentence seems dubious to me. $\endgroup$
    – Alec Rhea
    May 4, 2023 at 2:27
  • $\begingroup$ math.stackexchange.com/a/1619246/622 $\endgroup$
    – Asaf Karagila
    May 4, 2023 at 4:42
  • $\begingroup$ @JamesHanson Yep $\endgroup$ May 4, 2023 at 19:24
  • $\begingroup$ @AlecRhea Does that include proper classes definable from set parameters? $\endgroup$ May 4, 2023 at 19:25

1 Answer 1

10
$\begingroup$

Yes. If you start with infinitely many urelements and then take the sets whose kernel is finite (the kernel of a set is the set of urelements in its transitive closure), the resultant inner model will satisfy Replacement. A more generalized argument is included in my dissertation (https://arxiv.org/abs/2303.14274, Theorem 26).

Let ZFU$_\text{R}$ denote ZF set theory with urelements axiomatized with Replacement. In general, given a model $U$ of ZFU$_R$ an ideal $\mathcal{I}$ of the class of all urelements that contains every urelement singleton, the inner model $U^\mathcal{I}$, which contains all urelements and sets whose kernel is in $\mathcal{I}$, is a model of ZFU$_R$ also. And $U^\mathcal{I}$ satisfies AC if $U$ does. So in the case of ZFCU$_\text{R}$, the Hartogs number of a proper class can be any infinite cardinal.

For ZFCU$_\text{R}$ + Collection, the Hartogs number of a proper class cannot be any limit cardinal by the argument you have given. But this is the only constraint: for every successor infinite cardinal $\kappa^+$, there can be models of ZFCU$_\text{R}$ + Collection where the Hartogs number of the class of urelements is $\kappa^+$. This follows from Lemma 21 and Theorem 26 in my dissertation.

The result for ZFCU$_\text{R}$ can be extended to GBU$_\text{R}$ (in fact, KMU$_\text{R}$) with a global choice function. This is due to Felgner (see Theorem 108 in my dissertation). In class theory with urelements, it is important to distinguish different versions of second-order AC. For example, the existence of a global choice function doesn’t imply the universe can be well-ordered, as shown in Felgner’s model.

However, it is not known, for example, if KMU$_\text{R}$ + Collection + Global Choice (not Global Well-Ordering) is consistent with a proper class with Hartogs number, say, $\aleph_1$.

$\endgroup$
9
  • $\begingroup$ I think in your dissertation you also have the case of a proper class with Hartog number $\aleph_1$, and indeed this example has the urelement theory with collection and not merely replacement. Isn't it true that one can arrange such a situation with any given uncountable cardinal $\kappa$? (Or at least successor cardinals?) Can we have such classes for more than one $\kappa$ at a time, or perhaps even all $\kappa$? $\endgroup$ May 4, 2023 at 11:51
  • $\begingroup$ @JoelDavidHamkins I've edited my answer to address your comment. $\endgroup$
    – Bokai Yao
    May 4, 2023 at 15:39
  • $\begingroup$ Oh, cool! So global choice is weaker than limitation of size. This raises the question of whether the Hartogs number of a proper class can be $\aleph_1$ even without urelements. $\endgroup$ May 4, 2023 at 19:21
  • $\begingroup$ $\aleph_0$ is out because replacement implies collection in the absence of a proper class of urelements. $\endgroup$ May 4, 2023 at 19:28
  • $\begingroup$ Oh, that's a shame. So there's no model of this in ZF. Well, that answers all my questions on the topic thoroughly: the class of atoms can have any Hartogs number that's an infinite successor cardinal, and with replacement instead of collection it can also be an infinite limit cardinal. I'm glad to see someone else has asked the question. $\endgroup$ May 4, 2023 at 23:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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