This is a question about ZFC (or maybe NGB), but it is motivated by Randall Holmes' article on alternative set theories, especially his elaborations on TST, bounded Zermelo set theory and pocket set theory. Pocket set theory seems to derive its strength from the limitation of size axiom:

(limitation of size) – A class is a proper class if and only if it is equinumerous with all proper classes.

I read that NBG also verifies limitation of size, and that NBG + the generalized continuum hypothesis (GCH) is equiconsistent with ZFC. I don't really like the generalized continuum hypothesis, but it seems to be an adequate way to express the analog of "limitation of size" (and some intuitions about a size based set theory) for "sub-universes" of NGB.

Question: Is bounded Zermelo set theory + limitation of size + GCH identical to NBG + GCH?

This question is slightly ill posed, because bounded Zermelo set theory doesn't have proper classes, hence it is unclear what is meant by adding limitation of size. Maybe a "better" question would be

Question: Is bounded Zermelo set theory + GCH identical to ZFC + GCH

Bonus question: What can be said about the consistency strength of pocket set theory?

The idea behind this question is that bounded Zermelo set theory is equiconsistent with the simple theory of types TST (or typed set theory), and TST essentially embodies the philosophical position of Frank P. Ramsey and Rudolf Carnap, who accepted the ban on explicit circularity, but argued against the ban on circular quantification. This gives a very good reason to expect that bounded Zermelo set theory is consistent, but how does this relate to ZFC? The "limitation of size" principle seems (to me) like an impredicative principle with the potential to lead to a set theory equiconsistent with ZFC. The formulation via GCH is "ugly", hence I felt the need for these excessive explanations and motivations.

  • $\begingroup$ This question is based on a wrong assumption about the reason for the strength of pocket set theory (not limitation of size, but an impredicative class comprehension axiom leads to the strength). If this wrong assumption is fixed, all that remains of the question is basically which axioms can be omitted from NBG, if "limitation of size" (and maybe also GCH) is added as an axiom. The answer to that question can probably be found quite easily, potentially even on wikipedia... $\endgroup$ Jun 23, 2014 at 0:15
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    $\begingroup$ It is straightforward to show that limitation of size is equivalent to (second-order) Replacement and Global Choice over bounded Z + predicative second-order comprehension. Let T = bounded Z + LoS + predicative SO-comp. Then it follows that T + GCH is equivalent to NBG + Global Choice + GCH. Assuming NBG doesn't include Global Choice already, I take it that NBG + GCH won't imply T + GCH. However, since NBG + Global Choice is conservative over NBG + Choice and since GCH implies Choice, T + GCH is conservative over NBG + GCH. $\endgroup$ Jun 23, 2014 at 7:01
  • $\begingroup$ Just so you know, the Axiom of Limitation of Size implies, with only a little axiomatization (specifically $\mathrm{Ord}$ is well-ordered by $\in$ and every subclass of a well-orderable class is well-orderable), that every class is well-orderable. This is a strengthening of AC, which was already a strong axiom to begin with. ZF + Axiom of Limitation of Size is strong enough to prove ZFC. $\endgroup$
    – Zetapology
    Oct 12, 2017 at 1:56

2 Answers 2


I might be missing something, but if $V$ is a model of $\sf ZFC$, then $V_{\omega+\omega}$ is a model of Zermelo's set theory (and then some, since it also has regularity).

If $V\models\sf GCH$ then $V_{\omega+\omega}\models\sf GCH$ as well. So the answer is negative.

Allow me to propose a natural suggestion for a limitation axiom schema: If $\varphi(x,p)$ is a formula such that whenever we fix $p$ there is no set $y$ such that $u\in y\leftrightarrow\varphi(u,p)$, then there is a definable surjection from $\{u\mid\varphi(u,p)\}$ onto the universe.

Edit: As Yair points out, this schema ultimately quantifies over formulas as well, so it's no good for a first-order variant. Instead let me propose the following.

If $X$ is a proper class, then for every set $u$ in the universe, and every subset $x$ of $X$, there is a subset $y$ of $X$ such that $x\subseteq y$ and $y$ can be mapped onto $u$.

This says, effectively, that if $X$ is a proper class, then every "small" part of it can be extended to cover any small part of the universe. This means that there is a definable directed system of partial functions whose limit is a surjection from $X$ onto the universe.

In the presence of global choice, anyway, it seems that it should be possible to prove the existence of this limit (and if we require that all the functions in the system are injective, then we end up with a bijection too).

End of edit.

We can notice now that $V_{\omega+\omega}$ doesn't satisfy a reasonable schema of limitation of size (namely, for every formula, it defines a set if and only if there is no definable bijection with the universe). In particular because we have that $\{\omega+n\mid n\in\omega\}$ is a definable class which is not a set.

As for the consistency of Pocket set theory, here's something to contribute. Suppose that $V\models\sf ZFC+CH$. Then $H(\aleph_1)$, the set of all hereditarily countable sets, has size $\aleph_1$. If we consider its power set as the universe of Pocket set theory (countable sets are sets, uncountable sets are classes), then it should be okay.

  • $\begingroup$ I guess you mean $x\in y\leftrightarrow\varphi(x,p)$ instead of $u\in y\leftrightarrow\varphi(y,p)$. A definable surjection onto the universe sounds a bit too strong. At least I think that "limitation of size" is meant in a weaker sense than "definable", just like the axiom of choice also doesn't require a "definable" choice function. $\endgroup$ Jun 22, 2014 at 23:31
  • $\begingroup$ Well, the axiom of choice asserts something about sets which are objects. If you want to lose definability then you need to add classes to your language. $\endgroup$
    – Asaf Karagila
    Jun 22, 2014 at 23:41
  • $\begingroup$ Yes, I have to add classes to my language, and I even have to renounce on bounded quantification. As your counterexample shows, I didn't pay close enough attention that (A2) (class comprehension) of pocket set theory uses circular quantification in the way allowed by the cited philosophical position of Ramsey and Carnap. Looks like the consistency strength of pocket set theory has a quite "mundane" explanation, and that "limitation of size" is closer related to choice (and maybe replacement) than to "raw" consistency strength. $\endgroup$ Jun 23, 2014 at 0:00
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    $\begingroup$ @AsafKaragila: Your axiom schema still quantify over "definable surjections" - which are formulas. Maybe you want to use the rank function or something similar in order to get a specific surjection for each $\varphi$. $\endgroup$
    – Yair Hayut
    Jun 23, 2014 at 4:34
  • $\begingroup$ @Yair: You're right. In my defense, I did say that it is a naive suggestion! $\endgroup$
    – Asaf Karagila
    Jun 23, 2014 at 12:58

Question: Is bounded Zermelo set theory + limitation of size + GCH identical to NBG + GCH?

Sam Roberts suggested in a comment to add also (proper classes and) predicative second-order comprehension in order to allow a reasonable interpretation of this question. The resulting system is equiconsistent with NBG + GCH, but the axiom of foundation cannot be proved in it. Hence it is not equivalent to NBG + GCH, i.e. the comment is not fully accurate. However, this is just another oversight of the question itself (bounded Zermelo set theory is missing foundation), and the comment should be interpreted as an answer to the "intended" question, not to the actual "buggy" question.

Question: Is bounded Zermelo set theory + GCH identical to ZFC + GCH

Definitively not. Asaf Karagila's answer shows that it is not even equiconsistent.

This question is based on a wrong assumption about the reason for the strength of pocket set theory (not limitation of size, but an impredicative class comprehension axiom leads to the strength).

After reading Sam Roberts comment, I realized that an impredicative class comprehension axiom alone is not enough for a high consistency strength. One also needs axioms connecting the second order variables with the first order variables, like the induction axiom in Peano arithmetic. Possible axioms are induction, separation, replacement and collection (see Kripke Platek set theory for induction and collection). They should probably be preferred, because they don't mix global choice and consistency strength, and don't suggest a connection to the generalized continuum hypothesis which isn't there.

The axiom of induction is weak in terms of consistency strength, but it is independent of the other axioms (it is a variant of the axiom of foundation). Both replacement and collection should lead to the same consistency strength as limitation of size, without implying global choice. The axiom of separation is weaker in terms of consistency strength, and is implied by replacement.

Bonus question: What can be said about the consistency strength of pocket set theory?

It should be stronger than (impredicative) second order arithmetic. It is probably equiconsistent with (impredicative) third order arithmetic. It should be weaker than TST, which is equiconsistent to bounded Zermelo. So it should be possible to prove its consistency inside bounded Zermelo, by showing that the class of hereditarily countable sets is a set (i.e. exists as a set in bounded Zermelo) and is a model of pocket set theory.


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