Difference between revisions of "Reinhardt"

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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:LargeCardinalsBeyondChoice</cite>
 
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:LargeCardinalsBeyondChoice</cite>
  
Totally Reinhardt cardinals are the ultimate conclusion of the Vopěnka hierarchy. A cardinal is $n-$fold Vopěnka  if and only if, for every $A\subseteq V_\kappa$, there is some $\alpha\lt\kappa$ $\eta-$fold extendible for $A$ for every \(\eta\lt\kappa\), in that the witnessing embeddings fix $A\cap V_{\zeta_i}$. 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 $\eta$.
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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==
 
==Relations==

Revision as of 19:13, 21 August 2019

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.

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).[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 $V_\kappa\prec V_\lambda\prec V_\gamma$ (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 $\kappa$ is called $A$-super Reinhardt if for all ordinals $\lambda$ there is a non-trivial elementary embedding $j : V \rightarrow V$ such that $\mathrm{crit}(j) = \lambda$, $j(\kappa)\gt\lambda$ and $j^+(A)=A$. (where $j^+(A) := \cup_{α∈\mathrm{Ord}} j(A ∩ V_α)$)[3]

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”}$.[3]

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 $I0$.[1]

$\mathrm{WRA}$ (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.[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)”}\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 $(V_\gamma , V_{\gamma+1})\vDash \mathrm{ZF}_2 + \text{“There is a Reinhardt cardinal”}$.[3]

If $\delta_0$ is the least 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 $I3$). [3] Each club Berkeley cardinal is totally Reinhardt.[3]

References

  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 Gödel and the Foundations of Mathematics: Horizons of Truth. Cambridge University Press, 2011. www   bibtex
  3. Bagaria, Joan. Large Cardinals beyond Choice. , 2017. www   bibtex
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