Is it possible to iterate elementary embeddability and reflect on those stages that are elementary embeddable to themselves?

The following is a formal capture of that idea:

To the language of $\sf ZF$ (i.e., mono-sorted $\sf FOL(=,\in)$) add primitive partial unary functions $W$ and $j$.

To the axioms of $\sf ZF$, add the following axioms:

Restriction: $\forall \alpha: W_\alpha \lor j_\alpha \to \operatorname {ordinal}(\alpha)$

Injectivity: $W_\alpha \land W_\beta \land \alpha \neq \beta \to W_\alpha \neq W_\beta$

Cumulation: $\forall \operatorname {ordinal} \alpha \exists \lambda: W_\alpha=V_\lambda$

Elementarity: $\forall \operatorname {ordinal} \alpha \, (j_\alpha: W_\alpha \to W_\alpha \land \\ \forall \vec{x} \in W_\alpha [ \phi(\vec x) \leftrightarrow \phi(j_\alpha[\vec x ])] \land \\ \exists x: j_\alpha (x) \neq x) \\\text {where } \phi \text { is purely set theoretic }$

Reflection: $\forall \vec{x} \in W_\alpha \, (\phi \to \phi^{W_\alpha})$

if $\phi$ [in Reflection] is a formula of the language of set theory + $``j_\alpha \!"$, meaning that $W$ doesn't occur in it and every occurence of $j$ must be subscripted with $\alpha$; also $``\alpha \!"$ only appears in $\phi$ as a subscript of $j$.

Where $V_\lambda$ stands for the $\lambda^{th}$ stage of the cumulative hierarchy, defined in the customary manner. $\phi^X$ stands for relativising all quantifiers in $\phi$ to $``\in X\!"$.

Is the above theory consistent relative to some large cardinal property? If so, Which one?

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    $\begingroup$ I believe several people already pointed it out, but the fact that your notation is non-standard undermines the understandability of your question. You said that $W_\alpha$ is a partial function, then what is the meaning of $W_\alpha$ as a prime formula? (Do you mean $\alpha$ is in a domain of $W$?) $\endgroup$
    – Hanul Jeon
    Commented Jan 18, 2023 at 18:53
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    $\begingroup$ Also, is there any restriction on $\phi$ in your formulation of Elementarity and Reflection? Formulation of the Elementarity of a Reinhardt embedding usually only considers formulas in which $j$ does not appear. $\endgroup$
    – Hanul Jeon
    Commented Jan 18, 2023 at 18:58
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    $\begingroup$ @HanulJeon, yes it means $\alpha$ is in the domain of $W$, it actually means $W(\alpha)$ but I don't like writing it this way. $\phi$ in elementarity axiom does't use $j$. In Reflection axiom, it can use it. $\endgroup$ Commented Jan 18, 2023 at 19:03
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    $\begingroup$ You may not like using $W(\alpha)$ instead of $W_\alpha$, yes, that is okay, but could you write "$W_\alpha$ is well-defined" instead of just putting $W_\alpha$ alone? $\endgroup$
    – Hanul Jeon
    Commented Jan 18, 2023 at 19:47
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    $\begingroup$ It was most definitely not obvious to me what you meant with that. Anyway, doesn't cumulation axiom imply that $W_\alpha$ exists for all $\alpha$? Or is it only meant to apply for $\alpha$ for which $W_\alpha$ is well defined? $\endgroup$
    – Wojowu
    Commented Jan 18, 2023 at 20:02

2 Answers 2


$\newcommand\Ord{\mathit{Ord}}$The edit to the question has changed it enough so I think it deserves its own answer.

I assume that in reflection we have $\forall \vec{x} \in W_\alpha \, (\phi \to \phi^{W_\alpha})$ holds for all ordinals $\alpha$.

The theory with this strong reflection is inconsistent in a strong sense.

Specifically it fails even when you only consider $W_0$.

Indeed assume that $W_0$ reflects all formulae in the language of $\{\in,j_0\}$ then it reflects $\eta=\exists x(x\notin \operatorname{Dom}(j_0))$, so let $p\in W_0$ be witness of $\eta^{W_0}$.

$W_0$ also reflects $\psi(y)=\exists x\in \Ord (y\in V_x)$, let $r$ be witness of $\psi(p)$ in $W_0$.

Lastly reflect $\phi(y)=y\in \Ord\land V_y\subseteq \operatorname{Dom}(j_0)$ to get that $W_0$ thinks that $\phi(r)$, but this is a contradiction to the choice of $r$, $p$ (note that I only chose 1 arbitrary element, $p$, so I didn't use any choice).

  • $\begingroup$ Are you using absolutness of "..is an ordinal" over transitive models of ZF in the last step, because $r$ is seen by $W_0$ as ordinal, it need not be so externally unless you are using some kind of absolutness of ordinality over $W_0$. $\endgroup$ Commented Jan 19, 2023 at 16:23
  • $\begingroup$ @ZuhairAl-Johar I am using the fact that "being an ordinal" is absolute between transitive models, you don't need them to be models of ZF ("$x$ is an ordinal" is $\Delta_0$), but yes, I did use that fact. With a bit more effort you can show that even if we don't assume $W_0$ is transitive your strong reflection principle implies "being is ordinal" is absolute between $V$ and $W_0$ $\endgroup$
    – Holo
    Commented Jan 19, 2023 at 16:37
  • $\begingroup$ Thanks! That's what I was searching for. $\endgroup$ Commented Jan 19, 2023 at 16:52

Any model of $ZF$+stationary proper class of $I3$ ordinals (whose consistency strength is at most $ZF$+$I2$) where $W_α$ is listing of $V_{κ}$ where $κ$ is $I3$ and $j_α$ the witness of $I3(V_κ)$ will be a model of your theory.

Injectivity and Cumulation are trivial, $ZF$+Elementarity is part of our assumption.

For Reflection, let $φ$ be any formula, by Levy's reflection principle there is a proper class club $C_φ$ that reflect $φ$ (note that Levy's reflection works even with a modified language).

Because there is a stationary set of $I3$ cardinals, there exists a $I3$ cardinal in $C_φ$, then this ordinal will witness Reflection of $φ$.

This section is a follow-up based on the comments.

Let $\cal L$ be the language of our theory (or any other language extending $\{∈\}$), and let $ZF_{ \cal L}$ be $ZF$ with schema over the language $\cal L$ then any model of $ZF$ is (under a suitable interpretations) a model of $ZF_{ \cal L}$.

The reason for this is that we do not have any axioms to restrict any of our new symbols, let interpret each relation symbol in $\cal L\setminus\{∈\}$ as a tautology, each function symbol as the identity, and each constant as the emptyset.

Because all 3 of those are definable, any formula containing any symbols in $\cal L$ will be equivalent to a formula in $\{∈\}$.

So $\forall \alpha \,( W_\alpha \models \sf ZF_{\cal L})$ doesn't imply $W_\alpha$ is a model of ZF+Reinhardt, because while it does satisfy schema over the symbol $j$ and $W$, it does not see any nee from the universe to itself (The symbol of $j$ in $W_α$ has no relation to the symbol $j$ in $V$)

Now what about the reflection principle? In our language we have $2$ extra symbols, $j,W$.

We can restrict ourselves to $α$ such that $W_α$ reflects enough to prevent the triviality we had in the second paragraph of the follow-up, but it doesn't matter.

To show that it doesn't matter, look at the sentence $∀x(x∈Ord\implies ∃y∈Ord(\text{such that $W_x=V_y$ and $j_x$ is nee from $W_x$ to itself}))$, assume it is reflected into $W_α$, why is $W_α$ not Reinhardt? The reason is that $W_α$ does not see all of the ordinals, so while we may have $α∈W_α$ (well, for this specific sentence if it is reflected to $W_\alpha$ then $V_\alpha=W_\alpha$ so $\alpha\notin W_\alpha$, but this is irrelevant to my point), the value of $W_α^{W_α}$ is a proper subset of $W_α$, so $j,W$ inside of $W_α$ cannot say anything about the real $W_α,j_α$, i.e. we can not see any nee on $W_α$ inside of $W_α$.

So how can we get Reinhardt? To get Reinhardt we must adjust the nee $j_α$ to $W_α$, define for a set $X,Y$ the model $(X,∈,Y^{(k)})$ be the model $(X,∈,Y)$ but the symbol for $Y$ is $k$ and add the following axiom (remember that as long as we don't add interpretations/axioms on new symbols, they don't change the theory, so we can require just $ZF_{\{∈,k\}}$ instead of $ZF_{\{∈,k,j,W\}}$):

For every ordinal $α$, $(W_{α},∈,j_{α}^{(k)})⊨ZF_{\{∈,k\}}$

This is exactly your theory with the extra axiom that "$W_α$ can see the nee on itself".

This theory is very strong, I want to say that this theory is equivalent to "ZF + there is stationary proper class of $α$ such that $V_α$ is a model of Reinhardt", but I am not 100% sure about the $⇒$ direction (the proof of $⇐$ direction is essentially the same as my original answer but replacing $I3$ with Reinhardt).

  • $\begingroup$ I thought this theory is stronger than Reinhardt cardinals $\endgroup$ Commented Jan 18, 2023 at 17:56
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    $\begingroup$ @ZuhairAl-Johar No, it is much weaker. The problem is that having a lot of elementary embedding doesn't give us a way to "stich them together" to get Reinhardt. And even if you had a way to stich them together there is no reason to believe that the "resulting stich" won't be the identity. (For example, it is consistent [relative to blah blah] that there is proper class of $I3$ ordinals, and for every ordinal $α$, there is some $β$ such that the elementary embedding for the $I3$ ordinals above $β$ are the identity on $V_α$, so the embeddings are "eventually the identity") $\endgroup$
    – Holo
    Commented Jan 18, 2023 at 18:06
  • $\begingroup$ If one adds the axoim that $\forall \alpha \,( W_\alpha \models \sf ZF)$, then would that interpret $\sf ZF + Reinhardt $ at each $W_\alpha$? $\endgroup$ Commented Jan 18, 2023 at 18:40
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    $\begingroup$ @ZuhairAl-Johar ZFC + I3 at $\alpha$ implies $V_\alpha\models$ ZFC, but $(V_\alpha,j)\not\models$ ZFC$_j$ where $j:V_\alpha\to V_\alpha$ witnesses I3 at $\alpha$. $\endgroup$
    – Farmer S
    Commented Jan 18, 2023 at 19:47
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    $\begingroup$ I supposedly assume it is an elementary embedding over $W_\alpha$. It is my fault that I did not list every axiom I should state to avoid your confusion. $\endgroup$
    – Hanul Jeon
    Commented Jan 18, 2023 at 20:12

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