4
$\begingroup$

This question is an extension of something I asked earlier here: Ordering of large cardinals by cardinality

I have seen large cardinals ordered by consistency strength in several places but no ordering by cardinality of the least instance, although this is probably common knowledge among experts. Based on information scattered in various articles and the answer to my prior question from Dr. Hamkins, here is what I have gathered so far (I have deliberately avoided "identity crisis" cases):-

Strongly inaccessible < Mahlo < Weakly Compact < Totally Indescribable < Measurable < Huge (1-Huge upwards) < Rank-into-Rank < Supercompact < Extendible

My questions are:

1) Is this ordering correct ? Maybe < should be replaced by $\leq$ in some cases ?

2) Any important types I have left out which can be placed in the ordering without ambiguity ?

3) If a Reinhardt cardinal existed (maybe in ZF), would it be at the top of this size hierarchy ? My intuition is that it should, since a Reinhardt cardinal is an "ultimate extendible cardinal" in a sense, but I find my intuition isn't terribly good on these issues.

$\endgroup$
4
$\begingroup$

I added some information elsewhere, but it was to long to put in one post: http://metaordinals.azurewebsites.net/?p=261

Part 1: Your ordering is correct, but the size of the gaps are much more complex than just $LC_1<LC_2$ can express, and that list might be misleading. There are two main types of limits. Stationary limits are classes of cardinals $LC_1$ such that for such cardinal $\kappa$, the set $\{\lambda<\kappa|\lambda\,satisfies\,LC_2\}$ is stationary in $\kappa$; and measurable limits, where there is a normal measure $D$ on $\kappa$ such that the set $\{\lambda<\kappa|\lambda\,satisfies\,LC_2\}\in D$. We can codify these into operators; the stationary limit operator $M(X)$, and the measurable limit operator $o(X)$. From here we get two degrees of transcendence. We can define stationary transcendence as the existence of a $\kappa-$complete normal filter $F$ on $\kappa$ closed under $M(X)$, and measurable transcendence as the existence of a $\kappa-$complete normal filter $F$ on $\kappa$ closed under $o(X)$.

The set of Mahlo cardinals are precisely the set of stationarily transcendent cardinals over the inaccessibles. Every weakly compact cardinal is stationarily transcendent over the Mahlo cardinals.

In my opinion the standard definition of totally indescribable cardinals has been butchered. Define a cardinal $\kappa$ as $\lambda-$indescribable if and only if for every $S\subseteq V_\kappa$ and $\lambda$th order sentence $\phi$, there exists some $\alpha<\kappa$ such that $(V_\alpha,\in,S\cap V_\alpha)\vDash \phi \leftrightarrow (V_\kappa,\in,S\cap V_\kappa)\vDash \phi$. Then it is consistent for a cardinal $\kappa$ to be $\lambda-$indescribable for $\lambda>\kappa$, and we can call a cardinal totally-indescribable if it is $\lambda-$indescribable for every cardinal $\lambda$. Under the typical definition the totally indescribable cardinals would be the $\omega-$indescribable cardinals.

The $\lambda+1-$ indescribable cardinals are stationarily transcendent over the $\lambda-$indescribable cardinals, and every critical point of a non-trivial elementary embedding $j:M\rightarrow M$ is totally indescribable. If I am not mistaken, every weakly measurable cardinals is $\Sigma^2_1-$indescribable.

Theorem: Every measurable cardinal is a measurable limit of totally indescribable cardinals.

$Proof.$ Consider the normal measure given by $D=\{X\subseteq \kappa|\kappa\in j(X)\}$. Then $D$ is $\kappa-$complete normal ultrafilter. Note that for every $\alpha$th order formula $\phi$, its revitalization $\phi^{V_\kappa}$ is first order and moreover there is a truth predicate for $V_\kappa$. Also note that if $V_{j(\kappa)}\vDash \phi(S)$, then $V_\kappa\vDash \phi(S\cap V_\kappa)$ and so $j(\kappa)$ is totally indescribable. Moreover, the statement $\kappa$ is totally indescribable is $\Sigma_1$ and $M$ is an inner model and so $M\prec_{\Sigma_1}V$ and so $M\vDash(j(\kappa)\,is\,totally\,indescribable)$. Therefore $U\in D$, where $U=\{\lambda<\kappa|\lambda\,is\,totally\,indescribable\}$.■

Beyond that, however, the gaps are much larger, being instead measurable gaps. The $n+1-$ are measurably transcendent over the $n-$huge cardinals (By a simple alteration to the standard proof that they are measurable limits of $n-$huge cardinals).

Theorem: The rank-into-rank cardinals are measurably transcendent over the $\omega-$huge cardinals.

$Proof.$ Consider the filter generated by sets of the from $C\cap U_\omega$, where $U_\omega=\{\lambda<\kappa|\lambda\,is\,\omega-huge\}$, and $C$ is club in $\kappa_n$. It is clearly a normal filter, assuming $U$ is stationary, and similarly if $U$ is stationary closed under diagonal intersection. Now suppose we have a non-trivial elementary embedding $j: V_\lambda\rightarrow V_\lambda$, then $\kappa_n$ is $\omega-$huge$^1$ and so it remains to show $U$ is stationary in $\kappa_n$. Consider the normal measure, $D=\{X\subseteq \kappa_n|\kappa_n\in j(X)\}$. Then $\kappa_{n+1}$ is $\omega-$huge and so $U_\omega\in D$, where $U_\omega=\{\lambda<\kappa|\lambda\,is\,\omega-huge\}$. To see that it is closed under $o(X)$, observe $o(C\cap U)=o(C)\cap o(U)$, and by a similar argument such $o(U)=C\cap U$, where $C$ is the class of measurable limits of $U$, and it is clear $o(C)$ is closed. To see that it is unbounded, first observe that as $C$ is club, for every normal measure $D$ we have $C\in D$. Consider the normal measure $D=\{X\subseteq \kappa_n|\kappa_n\in j(X)\}$. As $M\vDash (There\,is\,a\,normal\,measure\,on\,\kappa_{n+1}\,that\,contains\,j(C))$, we have $o(C)\in D$.■

The gap between supercompact and rank-into-rank cardinals is also quite large.

Theorem: The supercompact cardinals are measurably transcendent over the rank-into-rank cardinals.

$Proof.$ Every supercompact cardinal $\kappa$ is $\Sigma_2-$reflecting, and the property "$\kappa_n$ is rank-into-rank," "There exists a "rank-into-rank embedding," are both $\Sigma_2$$^1$. Moreover, it is easy to see that the supercompact cardinals are measurably transcendent over the rank-into-rank cardinals.■

Lemma: The extendible cardinals are measurably transcendent over the rank-into-rank cardinals.

$Proof.$ Simply replace "There exists a rank-into-rank cardinal" with "There exists a supercompact cardinal" and use the fact that extendible cardinals are $\Sigma_3-$reflecting.■

So in conclusion, this list is technically correct, yet very misleading, as it gives the impression that the size gaps are around uniform.

Part 2: Yes, you omitted several very important cardinals. First, I am not sure if you omitted these because you considered them identity crises, but the weakly measurable cardinals are above the totally indescribable cardinals but below (Or equal, assuming the $GCH$) the measurable cardinals. Also the weakly indescribable cardinals are rather interesting; if $\kappa$ is weakly indescribable $L_\kappa\vDash ZFC+V=L$. In fact, this fact doesn't even require Choice.

The most tragic omission is definitely the worldly cardinals. A cardinal $\kappa$ is worldly if and only if $V_\kappa\vDash ZFC$. The worldly cardinals are the smallest type of large cardinal; if $\kappa$ is inaccessible there is a $\kappa-$complete filter $F$ on $\kappa$ closed under the worldly limit point operator (By a simple alteration to $^2$).

The pseudo-uplifting cardinals are weaker than the Mahlo cardinals, yet every pseudo-uplifting cardinal is a limit of inaccessibles, and the pseudo-uplifting cardinals are stationarily transcendent over the $\Sigma_3-$reflecting cardinals. Therefore, assuming the existence of a supercompact cardinal, the pseudo-uplifting cardinals are stationarily transcendent over the supercompact cardinals, and the uplifting cardinals are limits of pseudo-uplifting cardinals. I am not sure if you omitted because you considered them identity crises, yet they are still interesting. Beyond the totally indescribable cardinals are the strongly unfoldable cardinals.

Theorem: The strongly unfoldable cardinals are stationarily transcendent over the totally indescribable cardinals.

$Proof.$ Consider the filter generated by sets of the from $C\cap U$, where $U=\{\lambda<\kappa|\lambda\,is\,totally\,indescribable\}$, and $C$ is club in $\kappa$. It is clearly a normal filter, assuming $U$ is stationary, and similarly if $U$ is stationary closed under diagonal intersection. Now suppose we have a non-trivial elementary embedding $j: M\rightarrow N$, then $(V_\kappa,\in,S)\prec(V_{j(\kappa)},\in,j(S))$. Then $j(S)\cap V_\kappa=S$ and so it remains to show $U$ is stationary in $j(\kappa)$. By the same logic as before, $j(\kappa)$ is totally indescribable, and moreover, if $T$ is the higher order truth predicate for $V_{j(\kappa)}$ then $V_\kappa\prec _T V_{j(\kappa)}$. Then let $C$ be club. Then there exists some $\alpha<j(\kappa)$ such that $V_\alpha\vDash C\,is\,club$ and $V_\alpha\prec _T V_{j(\kappa)}$, and so $\alpha\in C$ and $\alpha$ is totally indescribable. To see that it is closed under $M(X)$, observe $M(C\cap U)=M(C)\cap M(U)$, and by a similar argument such $M(U)=C\cap U$, where $C$ is the class of stationary limits of $U$, and it is clear $M(C)$ is club.■

Beyond the strongly unfoldable are the strongly unfoldable of degree $\alpha$ cardinals. $\kappa$ is strongly unfoldable of degree $\alpha$ if and only if for every $S\in H_{\kappa^+}$ and cardinal $\lambda$ there is a $\kappa-$model $S\in M$ non-trivial elementary embedding $j: M\rightarrow N$ with $N$ transitive and critical point $\kappa$ such that $j(\kappa)>max\{\alpha,\lambda\}$, such that $V_\alpha\subset N$ and $N\vDash (\kappa\,is\,strongly\,unfoldable\,of\,degree\,\beta)$ for every $\beta<\alpha$. A cardinal is totally unfoldable if and only if it is strongly unfoldable of degree $\alpha$ for every ordinal $\alpha$.

Theorem: The unfoldable with degree $\alpha+1$ cardinals are stationarily transcendental over the unfoldable with degree $\alpha$ cardinals. The superstrongly unfoldable cardinals are stationarily transcendent over the totally unfoldable cardinals.

$Proof.$ Every cardinal $\kappa$ that is unfoldable with degree $\alpha+1$ is clearly unfoldable with degree $\alpha$. Consider the filter generated by sets of the from $C\cap U$, where $U=\{\lambda<\kappa|\lambda\,is\,totally\,indescribable\}$, and $C$ is club in $\kappa$. It is clearly a normal filter, assuming $U$ is stationary, and similarly if $U$ is stationary closed under diagonal intersection. Now suppose we have a non-trivial elementary embedding $j: M\rightarrow N$, then $(V_\kappa,\in,S\cap V_\kappa)\prec(V_{j(\kappa)},\in,S)$ and so it remains to show $U$ is stationary in $j(\kappa)$. Then $j(\kappa)$ is unfoldable with degree $\alpha$. Let $D=\{X\subseteq \kappa|\kappa\in j(X)\}$. Then if $S\in D$, $S$ is stationary, and $N\vDash (j(\kappa)\,is\,unfoldable\,with\,degree\,\alpha)$ and so $U\in D$. To see that it is closed under $M(X)$, observe $M(C\cap U)=M(C)\cap M(U)$, and by a similar argument such $M(U)=C\cap U$, where $C$ is the class of stationary limits of $U$, and it is clear $M(C)$ is club. For the rest, see $^3$.■

The supercompact cardinals are more complex than simply being smaller then extendibles.

Theorem: If $\eta$ is not an infinite limit ordinal with $cf\eta<\kappa$, and $\kappa$ is $\eta+1-$extendible then $\kappa$ is $\beth_{\kappa+\eta}-$supercompact. If $\kappa$ is $\eta+2-$extendible then $\kappa$ is $\beth_{\kappa+\eta}-$supercompact. If $\eta<\kappa$ and $\kappa$ is $\kappa$ is $\beth_{\kappa+\eta}-$supercompact, then $\kappa$ is measurably transcendent over the $\eta-$extendible cardinals. Moreover, the $1-$extendible supercompacts are measurably transcendent over the supercompact cardinals.

$Proof.$ For the first part, see $^4$. For the second part, suppose $\eta$ is a limit ordinal with $cf\eta<\kappa$. Then $\xi=\eta+1$ is not a limit ordinal and so if $\kappa$ is $\xi-$extendible it is $\beth_{\kappa+\xi}-$supercompact. For the third part, let $j: V\rightarrow M$ witness the $|\beth_{\kappa+\eta}|-$supercompactness of $\kappa$, and let $j`=j\restriction V_{\kappa+\eta}$. Then $j': V_{\kappa+\eta}\rightarrow V_{j(\kappa+\eta)}$ is a non trivial elementary embedding witness the $\eta-$extendibility of $\kappa$. The rest follows simply. For the fourth part, let $j: V_{\kappa+1}\rightarrow V_{j(\kappa)+1}$ witness the $1-$extendibility of $\kappa$ and $D=\{X\subseteq \kappa|\kappa\in j(X)\}$. Then $V_{j(\kappa)}\vDash (\kappa\,is\,\lambda-supercompact\,for\,\lambda>\kappa\,and\,\lambda<j(\kappa))$. But $\kappa$ is supercompact . Therefore $U\in D$, where $U=\{\lambda<\kappa|\lambda\,is\,supercompact\}$. Then repeat a similar argument to the above.■

Most variants of huge cardinals are between the strongly unfoldable and superstrongly unfoldable cardinals. The almost huge cardinals are measurably transcendent over the Vopěnka cardinals. The $n+1-$huge cardinals are measurably transcendent over the almost $n+1-$huge cardinals, themselves measurably transcendent over the $n-$huge cardinals.

Finally, you missed the entire highest echelon of large cardinals. The superhuge cardinals are measurably transcendent over the extendible cardinals, and the super $n+1-$huge cardinals are measurably transcendent over the almost super $n+1-$huge cardinals, themselves measurably transcendent over the super $n-$huge cardinals. Beyond the $\omega-$superhuge cardinals are the stationarily superhuge cardinals, which obey a similar hierarchy. There are also the strong, superstrong, and Woodin cardinals.

Theorem: The $2-$strong cardinals are measurably transcendent over the measurable cardinals; moreover the $2^\kappa-$supercompact cardinals are measurably transcendent over the superstrong cardinals; moreover the $1-$extendible cardinals are measurably treanscendent over the superstrong cardinals. The Woodin cardinals are measurably transcendent over the superstrong cardinals.

$Proof.$ For all, a simple alteration to the proof in $^5$ will suffice.■

Part 3: They suffice, to an extent. Certainly they posses all the properties on your list, and some more. However, they are not larger than the superhuge cardinals, at least to my knowledge. The reason for this is that super Reinhardt cardinals are not trivial extensions of Reinhardt cardinals. However, every super Reinhardt cardinal is $\omega-$superhuge, and we can define stationarily Reinhardt cardinals that are stationarily $\omega-$superhuge.

References: $^1$ Upward reflection of rank-into-rank cardinals (Answer), Joel David Hamkins.

$^2$ https://kamerynblog.wordpress.com/2017/01/13/just-how-big-is-the-smallest-inaccessible-cardinal-anyway/, Kameryn Williams

$^3$ https://arxiv.org/pdf/1403.2788.pdf, Joel David Hamkins, Thomas A. Johnstone

$^4$ Extendibility vs supercompactness (Answer), Gabe Goldberq

$^5$ The Higher Infinite, 26. Extenders, Akihiro Kanamori

$\endgroup$
  • $\begingroup$ Thank you for the fantastic and detailed answer . Apologies for not checking back on this sooner to upvote. I think you should seriously consider writing this up as a short article ! $\endgroup$ – Anindya Jul 24 at 21:52
  • $\begingroup$ I'm glad I could help. I might write it up as you suggested. $\endgroup$ – Master Jul 26 at 22:27
  • $\begingroup$ Your definition of $\alpha$-indescribable uses the letter $\alpha$ for both the level of indescribability and the rank to which $\phi$ reflects. $\endgroup$ – Andreas Blass Jul 28 at 23:24
  • $\begingroup$ Thanks for fixing it so quickly! $\endgroup$ – Andreas Blass Jul 28 at 23:26
  • $\begingroup$ Thanks for catching that. I the indescribablity a little in general. $\endgroup$ – Master Jul 28 at 23:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.