# Can we interpret Reinhardt cardinals this way?

To the language of set theory add a primitive unary predicate $$\operatorname {Universe}$$ and a primitive total unary function $$j$$. Add all axioms of $$\sf ZF$$ in the language of this theory, i.e. the new primitives allowed to be used in instances of Separation and Replacement, I'll refer to this simply as $$\sf ZFj$$.

Cumulative: $$\forall X: \operatorname {Universe}(X) \to \exists \lambda: X=V_\lambda$$

Modeling: if $$\psi$$ is an axiom of $$\sf ZFj$$, and $$\psi^X$$ is the $$\in X"$$ bounded form of $$\psi$$; then: $$\forall X: \operatorname {Universe}(X) \to \psi^X$$

Elementarity: if $$\phi(x_1,..,x_n)$$ is a formula in signature $$\{=,\in\}$$, having $$x_1,..,x_n\! \!"$$ as its sole free variables, and none of them occur bound, then: $$\forall X: \operatorname {Universe}(X) \to\forall x \in X \, (j(x) \in X) \land \\\forall x_1,..,x_n \in X \\ (\phi(x_1,..,x_n) \iff \phi(j(x_1),..,j(x_n))) \\ \land \exists x \in X: j(x) \neq x$$

The intention is to render the restriction of $$j$$ to any Universe $$X$$ a non-trivial elementary embedding from $$X \to X$$.

Reflection: if $$\phi$$ is a formula [defined functions and predicates allowed] not using the symbol $$X$$, and $$\phi^X$$ is the $$\in X"$$ bounded form of $$\phi$$; then: $$\forall \vec{p} \, (\phi \to \exists X: \operatorname {Universe} (X) \land \phi^X)$$

Is this consistent? If so, what's its consistency strength?

In particular does it manage to interpret Reinhardt cardinals at each universe? Would we have a club of Reinhardt cardinals?

• Seems like a really overengineered way to state there is a lot of I3 cardinals. Commented Jan 20, 2023 at 8:52
• @AsafKaragila, why the universes do not manage to capture Reinhardt cardinals? Commented Jan 20, 2023 at 8:56
• You're just saying a lot $V_\lambda$ have elementary embeddings to itself. Commented Jan 20, 2023 at 9:03
• @AsafKaragila but $V_\lambda$ models $\sf ZFj$, that is $j$ is used in replacement. So, how is that different from the Reinhardt's cardinal situation. Commented Jan 20, 2023 at 9:06
• @FarmerS, let's take an example, take $\phi$ to be the formula $\exists q ( j(p)=q)$ then by reflection we'll have $(\exists X: \operatorname {Universe}(X) \land \exists q \in X ( j(p)=q))$ . I couldn't see the problem of non-well-definiteness? $j$ is a primitive symbol of the language. Can you please explain this point? Commented Jan 20, 2023 at 10:05

It is inconsistent. Call an ordinal b an independent critical point if for every for every Universe X,

if c∈X then there is a function f with domain X and an ordinal α, such that "f is an elementary

embedding from X to Vα" holds, for all x∈X f(x)∈X, and b is the least ordinal with f(b)≠b.

Note that the critical point of j is an independent critical point because the restriction of j to any

Universe X has the above property of f. Let c be the least independent critical point. By the

axiom schema of Reflection, there is a universe K such that "c is the least independent critical

point" holds relativized to K. By the definition of independent critical point, there is a function

g with domain K and ordinal γ such tha g is an elementary embedding from K to Vγ, and for all x∈K g(x)∈K

and c is the least ordinal with g(c)≠c. Let F(x) be a formula expressing

"x is the least independent critical point". Define a sequence s by

s0=c and s(n+1) is the least ordinal α such that α is greater than g(s(n)) and for K, Vα reflects all

subformulas of F. Let t=U{sn|n∈𝜔}. Then g(t)=t, and F(c) holds in V(t). By elementarity,F(g(c)) holds

in V(g(t)). That is F(g(c)) holds in V(t). But this is impossible.