9
$\begingroup$

Let $ZC$ be Zermelo set theory with choice, which differs from $ZFC$ in omitting the axiom scheme of replacement. EDIT: I think I want to include foundation in the axioms, which apparently isn't normally considered to be part of Zermelo set theory.

As is well-known, $ZC$ is much weaker than $ZFC$; for instance $V_{\aleph_\omega}$ models $ZC$. One way to measure the difference is by interpolating between the two with the theories $ZC_n$ for $n \in \mathbb N$, where one adds an axiom or axiom scheme of $\Sigma_n$ replacement; I believe one has $ZC = ZC_0$ and informally $ZFC = ZC_\omega$. At the extreme low end, I believe that $ZC$ doesn't even prove that every well-order is isomorphic to an ordinal, so if it makes things easier in the following to replace $ZC$ with $ZC_1$, I don't think I'd object.

What I'd like to know is whether $ZC$ admits a "large cardinal hierarchy" resembling in some sense the familiar hierachy of large cardinal axioms one can add to $ZFC$, and which give another way of calibrating consistency strength of such theories.

When trying to adapt large-cardinal ideas from $ZFC$ to $ZC$, I imagine there are plenty of potential issues.

  • For one thing, as usual equivalent formulations of a statement in $ZFC$ may become inequivalent in $ZC$, so one must think carefully about choosing "correct" formulations.

  • More concerningly, the canonical example of a model of $ZC$ given by $V_{\aleph_\omega}$ suggests that perhaps one way $ZC$ differs from $ZFC$ is that in $ZC$ the universe doesn't necessarily "extend endlessly upward", and so "adding to the top" by hypothesizing "large cardinals" may simply be an ineffective way to generate stronger extensions of $ZC$ which are not extensions of $ZFC$.

So I think the somewhat-more-specific questions I have are:

Question 1: Is there anything analogous to the large cardinal hierarchy when it comes to theories which extend $ZC$ but not $ZFC$?

Or is it rather the case that anything recognizable as a "large cardinal axiom" in $ZC$ will likely imply replacement anyway?

Question 2: If so, does this hierarchy lie strictly below $ZFC$ in consistency strength?

Or is it rather possible to get a theory without replacement which is stronger than $ZFC$ in consistency strength, or even incomparable?

Question 3: What's an example of an analog of a large cardinal axiom in $ZC$? (Or if the answer to (1) is "no", then: what are some other interesting ways to get theories between $ZC$ and $ZFC$ other than the theories $ZC_n$?)

My favorite large cardinal axiom happens to be Vopenka's principle. So for instance, is there a version of Vopenka in $ZC$, and if so, is the resulting theory weaker than $ZFC$?

I'd also be interested in asking similar questions about $BZC$, the material-set-theoretic analog of ETCS, where the language is modified so that there simply aren't any unbounded quantifiers at all. But perhaps that would be too radically different a setting from $ZFC$ to really get a grip on the question.

$\endgroup$
4
  • 1
    $\begingroup$ This is probably not what you had in mind, but just in case: if we weaken ZC even more and restrict Separation to Δ₀ formulae, but add Δ₀ replacement, we get Kripke-Platek (KP) + powerset. Now KP has its own hierarchy of “recursively large ordinals” (recursively inaccessible, recursively Mahlo, nonprojectible, etc.), which reflect the “small” large cardinals. The theory Power-KP obtained by adding a powerset function (rather than axiom) to KP has been studied, and seems to have much in common with KP (its transitive models are called “power-admissible” sets). … (contd.) $\endgroup$
    – Gro-Tsen
    Commented Feb 24, 2021 at 0:19
  • 1
    $\begingroup$ (contd.) … So I am tempted to believe that there should be notions like “power-recursively inaccessible”, “power-recursively Mahlo”, etc., waiting to be studied. AFAICT this hasn't received any attention, but it could be an interesting hierarchy to look at. $\endgroup$
    – Gro-Tsen
    Commented Feb 24, 2021 at 0:22
  • $\begingroup$ @Gro-Tsen Thanks, that's good to know. As is probably apparent, I don't really know enough set theory to say with confidence that ZC is the base theory I'm "really" interested in. Just to be sure I have it straight, we have that ZC interprets KP+Powerset and KP+Powerset interprets BZC=Mac Lane, but between KP and Mac Lane neither one "directly" interprets the other, right? $\endgroup$ Commented Feb 24, 2021 at 0:27
  • 1
    $\begingroup$ I'm afraid I easily get lost in the twisty maze of theories all alike, but the reference here is Mathias's paper “The Strength of Mac Lane Set Theory”. Let's just say I was mostly suggesting a look at the theory $\mathsf{KP}^{\mathcal{P}}$ (in the terminology of this paper), which is somewhat better behaved than $\mathsf{Z}$ and presumably admits an interesting hierarchy of “small large” cardinals (whose existence in $\mathsf{ZFC}$ would be probable). $\endgroup$
    – Gro-Tsen
    Commented Feb 24, 2021 at 10:55

2 Answers 2

6
$\begingroup$

Overall, the large cardinal axiom hierarchy is very similar between ZC (ZFC minus replacement; we are including regularity) and ZFC. A large cardinal axiom (unprovable in ZFC) satisfied by $κ$ typically implies in ZC that $V_κ$ satisfies ZFC + weaker large cardinal axioms. However, the axioms typically do not imply additional replacement above $κ$ even for $Σ^V_3$ axioms such as existence of a strong or a supercompact cardinal (and hence their strength is truncated accordingly); but an extendible or a proper class of strong cardinals gives $Σ_2$ replacement, and a proper class of extendibles gives $Σ_3$ replacement.

Equiconsistencies also typically carry over to ZC. For example, ZC + $L(ℝ)⊨\text{AD}$ is equiconsistent with ZC + $ω$ Woodin cardinals whose supremum exists.

However, there are some differences.

One is notational. ZC does not prove that $ω2=ω+ω$ exists as the transitive set. However, ZC interprets $Σ_1$ replacement, and we can either add $Σ_1$-replacement, or implicitly speak of codes for ordinals and other sets.

Defining HOD requires replacement. However, there are inequivalent first order definable approximations of OD in ZC, one of which is $∪_{V_κ \text{ exists}} \mathrm{OD}^{V_κ}$.

The lack of singular infinite cardinals creates some strength differences. For example, ZC + $∀κ \, (κ^+)^L < κ^+$ is equiconsistent with ZC rather than ZC + $0^\#$. Still, there are other covering properties whose violation has high strength in ZC, and nonexistence of an inner model with a Woodin cardinal should still imply that the core model exists. Also, ZC($j$) + “$j$ is a nontrivial elementary embedding $V→V$” (called Wholeness Axiom; it proves ZFC without replacement for $j$-formulas) is consistent relative to the $\mathrm{I}_3$ axiom, as opposed to the Kunen inconsistency in ZFC($j$) due to the axiom of choice and existence of $j^ω(\mathrm{crit}(j))$.

There are large cardinal axioms for ZC that are implied by ZFC. Borel determinacy is equivalent to $∀r∈ℝ \,∀α < ω_1 \, ∃β \, L_β(r) ⊨ \text{“} ω_α \text{ exists”}$. Also, the least $α<β$ with $\mathrm{Theory}(V_α)=\mathrm{Theory}(V_β)$ are strictly between $ω_1^L$ and $c^+$.

Bounded quantifier ZC (BZC, also called Mac Lane set theory) is useful for some equiconsistency statements, and as a minimal base theory. For example, BZC + a proper class of Woodin cardinals is conservative over $\text{Z}_2 + \text{PD}$ (second order arithmetic with projective determinacy). In turn, key theorems about universally Baire sets relying on a proper class of Woodin cardinals can in fact be proved in BZC + a proper class of Woodin cardinals (even though it does not prove that the set of all universally Baire sets of reals exists).

By reflection, for every consistent axiom A, ZC+A has lower consistency strength than ZFC+A, but this need not apply to schemas. For example, Vopěnka's principle over ZC (or just BZC) implies ZFC.

$\endgroup$
2
  • $\begingroup$ Thanks, there's a lot to dig into in this answer -- it really seems to cover all the bases! One lingering question -- for a large cardinal property $A(\kappa)$, $ZFC+\exists\kappa A(\kappa)$ typically proves that $V_\kappa$ models $ZFC+\exists \lambda B(\lambda)$ for certain weaker large cardinal properties $B$. And in your answer, you say that $ZF+\exists \kappa A(\kappa)$ typically proves that $V_\kappa$ models $ZFC+\exists \lambda C(\lambda)$ for certain weaker large cardinal properties $C$. Should I be thinking that $B = C$? Or is there rather sometimes a drop in strength from $B$ to $C$? $\endgroup$ Commented Feb 23, 2021 at 23:37
  • 1
    $\begingroup$ @TimCampion There is always a drop in strength, though it typically manifests only with contrived $B$. However, when the axiom is not $Σ_2$, there are natural examples: For example, ZC + "$κ$ is strong" does not prove that there is a $λ^{+ω}$-strong $λ$ in $V_κ$. $\endgroup$ Commented Feb 24, 2021 at 0:01
5
$\begingroup$

There is a simple example of a large cardinal axiom that extends $\sf ZC$ and is stronger than $\sf ZFC$ and yet it can violate $\sf ZFC$. Take for example the theory with the following axioms:

  1. Universes: every set belongs to a Grothendieck universe.

  2. Denumerability: No Grothendieck universe can have more than finitely many Grothendieck universes inside it

  3. Separation: as in $\sf ZC$.

Where a Grothendieck universe can be defined as an extensional well founded transitive set that is closed under powerset, set unions and not-greater in cardinality than operators.

This theory clearly violates $\sf ZFC$, yet it does interpret $\sf ZFC$, and even some versions of $\sf MK$

By the way the wholeness axiom can be considered as an extension of $\sf ZC$, since it doesn't really extend $\sf ZFC$ for all formulas of its language!

$\endgroup$
12
  • $\begingroup$ Thanks! Maybe I'm missing something obvious -- in what sense does this theory "violate" ZFC? I think I believe that the theory neither extends nor is extended by ZFC in the strict sense -- i.e. it proves things that ZFC does not prove, and vice versa. But if I understand correctly, one can add the scheme of replacement to this theory to arrive at an extension of ZFC which is consistent (provably relative to large cardinals). So I take it you don't mean that adding replacement to this theory results in inconsistency -- right? $\endgroup$ Commented Feb 14, 2021 at 20:05
  • 2
    $\begingroup$ this theory falsifies repalcement, since obviousely you cannot have a set of all Grothendieck unvierses here, but we only have countably many of those? If you add replacement to this theory, it becomes inconsistent! $\endgroup$ Commented Feb 14, 2021 at 20:12
  • 2
    $\begingroup$ @TimCampion (2) violates replacement since under replacement, we could take the supremum of the first $\omega$ inaccessibles. $\endgroup$ Commented Feb 14, 2021 at 20:13
  • 1
    $\begingroup$ @TimCampion, as far as I know, ZC is a very able theory as regards defining relations, It does capture the notion of a relation, and of cardinality very well, so I don't see why should it differ from the usual one. $\endgroup$ Commented Feb 14, 2021 at 20:25
  • 1
    $\begingroup$ @TimCampion, there are formulations of $ZC + rank$, Randall Holmes had spoken about this, and I think possibly Scott-Potter also alluded to that. You can easily have extensions of $ZC+rank$ that violate $ZFC$ and yet interpret it also. $\endgroup$ Commented Feb 14, 2021 at 20:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .