The existence of **Reinhardt cardinals** has been refuted in $\text{ZFC}_2$ and $\text{GBC}$ by Kunen ([](Kunen_inconsistency.md)), the term is used in the $\text{ZF}_2$ context, although some mathematicians suspect that they are inconsistent even there. ## Definitions 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 $V_\kappa\prec V$ ($\mathrm{WR}(\kappa)$. Existence of $\kappa$ is Weak Reinhardt Axiom ($\mathrm{WRA}$) by Woodin).{% cite Corazza2010 %}: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 $V_\kappa\prec V_\lambda\prec V_\gamma$ (for some $\gamma > \lambda > \kappa$).{% cite Baaz2011 %}:(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.{% cite Bagaria2017 %} 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.{% cite Bagaria2017 %} For a proper class $A$, cardinal $\kappa$ is called **$A$-super Reinhardt** if for all ordinals $\lambda$ there is a non-trivial elementary embedding $j : V \rightarrow V$ such that $\mathrm{crit}(j) = \kappa$, $j(\kappa)\gt\lambda$ and $j^+(A)=A$. (where $j^+(A) := \cup_{α∈\mathrm{Ord}} j(A ∩ V_α)$){% cite Bagaria2017 %} A **totally Reinhardt** cardinal is a cardinal $\kappa$ such that for each $A ∈ V_{κ+1}$, $(V_\kappa, V_{\kappa+1})\vDash \mathrm{ZF}_2 + \text{“There is an $A$-super Reinhardt cardinal”}$.{% cite Bagaria2017 %} Totally Reinhardt cardinals are the ultimate conclusion of the Vopěnka hierarchy. A cardinal is Vopěnka if and only if, for every $A\subseteq V_\kappa$, there is some $\alpha\lt\kappa$ $\eta-$extendible for $A$ for every \(\eta\lt\kappa\), in that the witnessing embeddings fix $A\cap V_\zeta$. In its original conception Reinhardt cardinals were thought of as ultimate extendible cardinals, because if $j: V\rightarrow V$ is elementary, then so is $j\restriction V_{\kappa+\eta}: V_{\kappa+\eta}\rightarrow V_{j(\kappa+\eta)}$. It is as if one embedding works for all $\eta$. ## Relations $\mathrm{WRA}$ (1) implies thet there are arbitrary large $I1$ and super $n$-huge cardinals. Kunen inconsistency does not apply to it. It is not known to imply $I0$.{% cite Corazza2010 %} $\mathrm{WRA}$ (1) does not need $j$ in the language. It however 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.{% cite Corazza2010 %} $\mathrm{WR}(\kappa)$ (1) implies that $\kappa$ is a measurable limit of [supercompact](Supercompact.md "Supercompact") cardinals and therefore is [](Strongly%20compact.md). It is not known whether $\kappa$ must be supercompact itself. Requiring it to be [extendible](Extendible.md "Extendible") makes the theory stronger.{% cite Corazza2010 %} Weakly Reinhardt cardinal(2) is inconsistent with $\mathrm{ZFC}$. $\mathrm{ZF} + \text{“There is a weakly Reinhardt cardinal(2)”}\rightarrow\mathrm{Con}(\mathrm{ZFC} + \text{“There is a proper class of $\omega$-huge cardinals”})$ (At least here $\omega$-huge=$I1$) (Woodin, 2009). You can get this by seeing that $V_\gamma\vDash\forall\alpha\lt\lambda(\exists\kappa'\gt\alpha(I1(\kappa')\land\kappa'\lt\lambda))$. If $\kappa$ is super Reinhardt, then there exists $\gamma\lt\kappa$ such that $(V_\gamma , V_{\gamma+1})\vDash \mathrm{ZF}_2 + \text{“There is a Reinhardt cardinal”}$.{% cite Bagaria2017 %} If $\delta_0$ is the least [Berkeley](Berkeley.md "Berkeley") cardinal, then there is $\gamma\lt\delta_0$ such that $(V_\gamma , V_{\gamma+1})\vDash\mathrm{ZF}_2+\text{“There is a Reinhardt cardinal witnessed by $j$ and an $\omega$-huge above $\kappa_\omega(j)”$}$. (Here $\omega-$huge means $I3$). {% cite Bagaria2017 %} Each club Berkeley cardinal is totally Reinhardt.{% cite Bagaria2017 %}