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A Reinhardt cardinal is defined as the critical point of a nontrivial elementary embedding $j: V\rightarrow V$ from the universe $V$ to itself, and is known to be inconsistent with the axiom of choice when paired with standard ZFC. Are Reinhardt cardinals still inconsistent with the axiom of choice when paired with the weaker BZC, Zermelo set theory with only bounded separation and choice?

Asaf Karagila
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1 Answers1

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Your theory is consistent modulo large cardinal axioms. You can see that if $\lambda$ witnesses $I_3$ (i.e., there is a non-trivial elementary embedding $j: V_\lambda\to V_\lambda$), then $V_\lambda$ satisfies the followings:

  1. $\mathsf{ZFC}$ (but separation and replacement schemas does not allow $j$ to appear,)
  2. The Wholeness axiom $\mathsf{WA}$ (separation schema for formulas with $j$.)

Hence $V_\lambda$ satisfies $\mathsf{BZC}$ with $j$ allowed to appear in bounded separation.

If you do not require $j$ can appear in bounded separation, then weaker large cardinal axioms suffice to provide the consistency of $\mathsf{BZC}$ with an elementary embedding $j:V\prec V$; for example, if $0^\sharp$ exists and $j:L\to L$ be an elementary embedding which sends the $n$th Silver indiscernibles to $(n+1)$th, and fixes all other indiscernibles, then $L_\iota$ for $\omega$th Silver indiscernible $\iota$ thinks it is a model of $\mathsf{ZFC}$ with an embedding $j:L_\iota\to L_\iota$.

Hanul Jeon
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