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This posting is related to this posting, and builds its motivation from this answer to it.

Define: $\operatorname {History}(x) \iff \\\forall y \in x: y=\{c \mid \exists z : z \in y \cap x \land (c \in z \lor c\subseteq z)\}$

Define: $\operatorname {Level}(l) \iff \exists h: \operatorname {History}(h) \land l \in h$

Axioms:

SETS: for each partial function $f$ that is definable after a functional relation, we have: $$\forall a \exists l \ni a : \operatorname {Level}(l) \land \\ \forall x \in l \exists!y \in l: y=\{f(z) \mid z \in x\}$$

From this single schema one can derive all axioms of $\sf ZF$.

For $\sf ZF$-$\sf Reg.$, we just replace $\operatorname {Level}(l)$ by $\operatorname {Supertransitive}(l)$, in the above formula.

I intend to get to $\sf ZFC$ along this line of thought. One possible way is to think of having Superlevel, which is a level as big as its history. Formally:

Define: $h(l) = \{x \in l \mid \operatorname {Level}(x)\}$

Define: $\operatorname {Superlevel}(l) \iff \operatorname {Level}(l) \land l \hookrightarrow h(l)$

Where "$\hookrightarrow$" means "is injective to".

Now, we replace $\operatorname {Level}(l)$ by $\operatorname {Superlevel}(l)$ in the formula of the schema of SETS.

Is it consistent to phrase the above formula in terms of Superlevels?

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  • $\begingroup$ You say you have $f$ as a symbol, but then of course it would be definable. I think you mean it comes from a definable functional relation, so that your axiom gives the replacement property. Since the levels are the $V_\alpha$s and the histories are $\{V_\alpha\mid \alpha<\lambda\}$, it seems to me that superlevels are just $V_\omega$ and $V_\gamma$ where $\gamma>\omega$ is a beth fixed point $\gamma=\beth_\gamma=|V_\gamma|$. Is the idea that you hope to gain insight by recasting terminology for these standard notions? $\endgroup$
    – Joel David Hamkins
    Commented Jul 5 at 12:19
  • $\begingroup$ @JoelDavidHamkins, this is related to the prior posting where levels are not supposed to be explicit. And if we can use the iterative conception to coin a single axiom. But, my main question here is about if it is consistent to have those beth fixed points indexed stages be closed on the replacement property the way written here. I think this is provable in ZFC by reflection and related absoluteness results. I wanted to be sure of that. $\endgroup$
    – Zuhair Al-Johar
    Commented Jul 5 at 14:30
  • $\begingroup$ I don't understand the question exactly. Of course not every beth fixed point is going to satisfy replacement. For example, the first beth fixed point $\gamma=\beth_\gamma$ is going to have a definable cofinal sequence: $n\mapsto \beth_{\beth_{\ddots_{\beth_0}}}$, nested to depth $n$. $\endgroup$
    – Joel David Hamkins
    Commented Jul 5 at 14:33
  • $\begingroup$ But also, the way you phrase the question about $f$, having it as a "symbol", it isn't clear whether you have just one function or want the scheme definable functions. Only the latter is going to give you replacement. The former will just be replacement for that one $f$. $\endgroup$
    – Joel David Hamkins
    Commented Jul 5 at 14:34
  • $\begingroup$ @JoelDavidHamkins, the terminology used here is I believe first due to Dana Scott. I think this is the simplest way to capture the level intuition and are more appropriate to be written at axiomatic levels than the commonly used definitions by recursion. Having a single axiom schema for set theory to me is a technical matter rather than intuitive one. $\endgroup$
    – Zuhair Al-Johar
    Commented Jul 5 at 14:35

1 Answer 1

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I won't engage with the level terminology, but I believe your question is answered by the following observation.

Theorem. ZFC is equiconsistent with the theory ZFC + there is a closed unbounded class $C$ of fully correct cardinals $\kappa$.

A cardinal $\kappa$ is fully correct if $\varphi(a)\iff V_\kappa\models\varphi(a)$ for every formula $\varphi$ in the language of set theory and $a\in V_\kappa$. This is expressible as a scheme in ZFC, not as a single notion, and so one must take care with the metatheoretic issues here. We can write $V_\kappa\prec V$ to denote that $\kappa$ is fully correct.

The right hand side of the theorem is understood as a scheme of statements asserting that every given $\varphi$ is absolute between any $V_\kappa$ and $V$.

Proof. This is an immediate consequence of the reflection theorem and the compactness theorem. Any finite part of the scheme involves only finitely many formulas, and for these there is a class club $C$ as desired by reflection. $\Box$

The $V_\kappa$ for $\kappa\in C$ in this theory will form sufficient so-called "superlevels" in your theory, and they will satisfy all instances of your $f$-formalism replacement axiom.

The scheme is sometimes denoted $V_\kappa\prec V$, as I mentioned, but notice that this doesn't mean that $V\models (V_\kappa\models\text{ZFC})$, since we only get each instance of ZFC being true in $V_\kappa$, and this is weaker than satisfying ZFC as a theory, since $V$ itself might be $\omega$-nonstandard.

Update. I've realized I may have misunderstood from our discussion in the comments that you want a superlevel that satisfies every instance of replacement, which is what my answer above achieves, but it seems to me now that your desired scheme (as much as I understand from your OP) requires only that every instance of replacement is true in some superlevel. (Note also that, although you mention ZF, there is an AC issue, since if $l$ is a superlevel, then from $l$ mapping injectively into $h(l)$ will imply it is well ordered, since the levels are well ordered.)

In this case, the superlevel version is simply equivalent to, a consequence of, the level version. The reason is that in ZFC, for every particular $\varphi$, by the reflection theorem there is a club $C$ of ordinals $\lambda$ such that $\varphi$ is absolute between $V_\lambda$ and $V$. And now, the point is that every class club will contain a beth fixed point $\lambda=\beth_\lambda$, since those also form a club, and so every $\varphi$ reflects to a beth fixed point, which is a superlevel.

So the superlevel formalism seems to add nothing at all.

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  • $\begingroup$ From within ZFC the superlevel formalism doesn't add anything, that I just came to know that, and this proves it consistent if the level formalism is consistent, which is , as you've shown here, relative to ZFC. But, for the purposes of this formalism superlevels are intorduced to enable us prove the axiom of choice since the levels alone cannot prove it. $\endgroup$
    – Zuhair Al-Johar
    Commented Jul 6 at 15:51
  • $\begingroup$ @ZuhairAl-Johar: to check I understand, your aim is to get Choice just by considering superlevels rather than levels in your axiom scheme, where a superlevel $l$ is a level which is well-orderable because there is an injection from $l$ to its history (i.e. to the set of levels less than $l$). I see how this will work. But, if that is the aim, why not just consider this definition: $\text{Level}(l) \Longleftrightarrow \exists h(\text{History}(h) \wedge l \in s \wedge l\text{ is well-orderable})$? $\endgroup$
    – Tim Button
    Commented Jul 7 at 21:10
  • $\begingroup$ @TimButton, I wanted to motivate choice after level theory, the concept of well-orderable as well as the concept of well-founded both belong to relation theory, the first specifically belongs to order theory. Notice, that the same abovementioned single axiom schema if we put supertransitive instead of level, then we can derive all of $\sf ZF-Reg.$ and I could have simply added "well-founded supertransitivity" instead of levels to get Foundation. But, by then I won't be getting it from supertransitivity theory for example. I wanted choice to follow from a level concept. $\endgroup$
    – Zuhair Al-Johar
    Commented Jul 8 at 5:32
  • $\begingroup$ @TimButton, continuation.., superlevels can be also called square levels because their width (cardinality) is equal to their height (cardinality of their history), those are level related notions, and so it is justified to think about them in level theory, so if we phrase the main axiom schema in connection to them instead of levels, then we get choice. So, there is a level based motivation for choice. However, it is clear that this motivation is strong in the sense that there is no real need for it from the perspective of level theory contemplation. $\endgroup$
    – Zuhair Al-Johar
    Commented Jul 8 at 8:29

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