From Cantor's Attic
Jump to: navigation, search

The existence of Reinhardt cardinals has been refuted in $\text{ZFC}_2$ and $\text{GBC}$ by Kunen (Kunen inconsistency), the term is used in the $\text{ZF}_2$ context, although some mathematicians suspect that they are inconsistent even there.


A weakly Reinhardt cardinal(1) is the critical point $\kappa$ of a nontrivial elementary embedding $j:V_{\lambda+1}\to V_{\lambda+1}$ such that $V_\kappa ≺ V$ ($\mathrm{WR}(\kappa)$. Existence of $\kappa$ is Weak Reinhardt Axiom ($\mathrm{WRA}$) by Woodin).[1]:p.58

A weakly Reinhardt cardinal(2) is the critical point $\kappa$ of a nontrivial elementary embedding $j:V_{\lambda+2}\to V_{\lambda+2}$ such that $V_\kappa ≺ V_\lambda ≺ V_\gamma$ (for some $\gamma > \lambda > \kappa$).[2]:(definition 20.6, p. 455)

A Reinhardt cardinal is the critical point of a nontrivial elementary embedding $j:V\to V$ of the set-theoretic universe to itself.[3]

A super Reinhardt cardinal $\kappa$, is a cardinal which is the critical point of elementary embeddings $j:V\to V$, with $j(\kappa)$ as large as desired.[3]

For a proper class $A$, cardinal $κ$ is called $A$-super Reinhardt if for all ordinals $λ$ there is a non-trivial elementary embedding $j : V → V$ such that $\mathrm{crit}(j) = κ$, $j(κ) > λ$ and $j(A) = A$. (where $j(A) := ⋃_{α∈\mathrm{OR}} j(A ∩ V_α)$)[3]

A totally Reinhardt cardinal is a cardinal $κ$ such that for each $A ∈ V_{κ+1}$, $\langle V_κ , V_{κ+1} \rangle \models \mathrm{ZF}_2 + \text{“There is an $A$-super Reinhardt cardinal”}$.[3]


  • $\mathrm{WRA}$ (1) implies thet there are arbitrary large $\mathrm{I}_1$ and super $n$-huge cardinals. Kunen inconsistency does not apply to it. It is not known to imply $\mathrm{I}_0$.[1]
  • $\mathrm{WRA}$ (1) does not need $j$ in the language. It hovewer requires another extension to the language of $\mathrm{ZFC}$, because otherwise there would be no weakly Reinhardt cardinals in $V$ because there are no weakly Reinhardt cardinals in $V_\kappa$ (if $\kappa$ is the least weakly Reinhardt) — obvious contradiction.[1]
  • $\mathrm{WR}(\kappa)$ (1) implies that $\kappa$ is a measurable limit of supercompact cardinals and therefore is strongly compact. It is not known whether $\kappa$ must be supercompact itself. Requiring it to be extendible makes the theory stronger.[1]
  • Weakly Reinhardt cardinal(2) is inconsistent with $\mathrm{ZFC}$. $\mathrm{ZF} + \text{“There is a weakly Reinhardt cardinal(2)”} \implies \mathrm{Con}(\mathrm{ZFC} + \text{“There is a proper class of $\omega$-huge cardinals”})$ (at least here $\omega$-huge=I1) (Woodin, 2009).[2]
  • If $κ$ is super Reinhardt, then there exists $γ < κ$ such that $\langle V_γ , V_{γ+1} \rangle \models \mathrm{ZF}_2 + \text{“There is a Reinhardt cardinal”}$.[3]
  • Totally Reinhardt cardinals are obviously stronger than super Reinhardt.[3]
  • If $δ_0$ is the least Berkeley cardinal, then there is $γ < δ_0$ s.t. $\langle V_γ , V_{γ+1} \rangle \models \mathrm{ZF}_2 + \text{“There is a Reinhardt cardinal witnessed by $j$ and an ω-huge above $κ_ω(j)”$}$.[3]
  • Each club Berkeley cardinal is totally Reinhardt.[3]


  1. Corazza, Paul. The Axiom of Infinity and transformations $j: V \to V$. Bulletin of Symbolic Logic 16(1):37--84, 2010. www   DOI   bibtex
  2. Baaz, M and Papadimitriou, CH and Putnam, HW and Scott, DS and Harper, CL. Kurt G{\"o}del and the Foundations of Mathematics: Horizons of Truth. Cambridge University Press, 2011. www   bibtex
  3. Bagaria, Joan. Large Cardinals beyond Choice. , 2017. www   bibtex
Main library