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# Large cardinals ordered by cardinality

Based on my answer to this question on Math Overflow:
https://mathoverflow.net/questions/250978/large-cardinals-ordered-by-cardinality-of-least-instance/333418#333418

1. The standard ordering of large cardinals.

The gaps between large cardinals are much more complex then just $$LC_1\lt LC_2$$ can express. There are two main types of large cardinals. Limit cardinals and embedding cardinals. A limit cardinals is created by taking a class of cardinals $$C$$, and asserting that $$\kappa\in F(C)$$ for some operator $$F$$. The three that will be used here is $$T(C)=\{\lambda\in C|cf\lambda\neq\omega\land\sup(C\cap\lambda)=\lambda\}$$ for club $$C$$, $$T'(C)=\{\lambda\in C|\sup(C\cap\lambda)=\lambda\}$$ for unbounded $$C$$, and $$M(C)=\{\lambda\in C|cf\lambda\neq\omega\land C\cap\lambda\text{ is stationary in }\lambda\}$$ for unbounded $$C$$.

Meanwhile, embedding cardinals are created asserted that there exists a non-trivial elementary embedding $$j:M\rightarrow N$$ such that $$N$$ satisfies some closure property $$P$$; e.g. $$N^\lambda\subseteq N$$ for some $$\lambda$$ or $$N^{j(\kappa)}\subseteq N$$, for $$\kappa$$ the critical point of $$j$$.

For each individual $$P$$, we define $$o_P(C)=\{\lambda\in C|\exists D(D\text{ is a normal measure generated by a }P\text{ embedding}\land C\cap\lambda\in D)\}$$ (It follows immediately $$D$$ is a normal measure on $$\lambda$$).

The limit of this process is asserting the exists of a $$\kappa-$$complete normal filter on $$\kappa$$, closed under $$F(C)$$. If we take $$F(C)=T'(C)$$, and assert $$\{\lambda\lt\kappa|\lambda\text{ is regular}\}$$ is in the filter, we get the Mahlo cardinals$$^1$$.

In my opinion the standard definition of totally indescribable cardinals has been butchered. Define a cardinal $$\kappa$$ as $$\alpha-$$indescribable if and only if for every $$\alpha$$th order formula $$\phi$$ and $$S\subseteq V_\kappa$$, there exists some $$\lambda\lt\kappa$$ such that $$(V_\lambda,\in,S\cap V_\lambda)\vDash\phi\leftrightarrow (V_\kappa,\in,S)\vDash\phi$$.

Theorem: If $$\kappa$$ is $$\alpha+1-$$indescribable, then there exists a $$\kappa-$$complete normal filter $$F$$ on $$\kappa$$, closed under $$M(C)$$, such that $$\{\lambda\lt\kappa|\lambda\text{ is }\alpha-\text{indescribable}\}\in F$$.

$$Proof.$$ Let $$F$$ be the $$\alpha+1-$$indescribable filter on $$\kappa$$. It is immediately normal and $$\kappa-$$complete. The assertion $$S$$ is stationary is $$\Pi_1^1$$, and so if $$S\in F$$ is stationary, then there is a class of $$\lambda\lt\kappa$$ such that $$S$$ is stationary in $$\lambda$$. Note that if $$C$$ is club, $$C\in F$$, and the rest follows from the fact that “$$\kappa$$ is $$\alpha-$$indescribable” is $$\alpha+1$$th order.■

Theorem: Every critical point $$\kappa\gt\alpha$$ of a non-trivial elementary embedding $$j:M\rightarrow M$$, such that $$M\vDash ZFC$$ is transitive, has a $$\kappa-$$complete normal filter $$F$$ on $$\kappa$$, closed under $$M(C)$$, such that $$\{\lambda\lt\kappa|\lambda\text{ is }\lt\lambda-\text{indescribable}\}\in F$$, and if $$V_{\kappa+\alpha}\subseteq M$$, then $$\kappa$$ is $$\alpha-$$indescribable.

$$Proof.$$ We first show $$M\vDash\kappa\text{ is }\lt\kappa-\text{indescribable}$$. Note that $$M\vDash((V_\kappa,\in,S\cap V_\kappa)\vDash\phi)\leftrightarrow M\vDash((V_{j(\kappa)},\in,S)\vDash\phi)$$. Then $$M\vDash(\exists\lambda\lt j(\kappa)(V_\lambda,\in,S\cap V_\lambda)\vDash\phi)$$, and so $$M\vDash(\exists\lambda\lt\kappa(V_\lambda,\in,S\cap V_\lambda)\vDash\phi)$$. Therefore $$M\vDash(\kappa\text{ is }\lt\kappa-\text{indescribable})$$.

I claim that $$F=\{X\subseteq\kappa|\kappa\in j(X)\}$$ is the necessary type of filter. It is immediate that it is $$\kappa-$$complete and normal. To see that it is closed under $$M(X)$$, note that if $$C$$ is club $$\kappa\in j(C)$$ and so $$X$$ is stationary below $$\kappa$$. Now, assume $$X\cap C\neq\emptyset$$. Then, as $$\kappa\subseteq M$$, $$X\cap C\cap M\neq\emptyset$$, and so stationarity is preserved in $$M$$, and so $$\kappa\in j(M(X))$$, and it is immediate given the above $$\{\lambda\lt\kappa|\lambda\text{ is }\alpha-\text{indescribable}\}\in F$$. If $$V_{\kappa+\alpha}\subseteq M$$, then $$M\vDash((V_\kappa,\in,S\cap V_\kappa)\vDash\phi)\leftrightarrow (V_\kappa,\in,S\cap V_\kappa)\vDash\phi$$.■

Theorem: If $$\kappa$$ is measurable, then there exists a normal measure $$D$$ on $$\kappa$$ such that $$\{\lambda\lt\kappa|\lambda\text{ is}\lt\lambda-\text{indescrbable}\}$$. If $$\kappa$$ is $$\lt\kappa-$$strong, $$\kappa$$ is $$\lt\kappa-$$indescrbable.

$$Proof.$$ Note that $$((V_\kappa,\in,S\cap V_\kappa)\vDash\phi)\leftrightarrow M\vDash((V_{j(\kappa)},\in,S)\vDash\phi)$$. Then $$M\vDash(\exists\lambda\lt j(\kappa)(V_\lambda,\in,S\cap V_\lambda)\vDash\phi)$$, and so $$(\exists\lambda\lt\kappa(V_\alpha,\in,S\cap V_\lambda)\vDash\phi)$$. Therefore $$M\vDash(\exists\lambda\lt\kappa(V_\lambda,\in,j(S\cap V_\lambda))\vDash\phi)$$. Therefore $$M\vDash(\kappa\text{ is }\lt\kappa-\text{indescribable})$$. The other part follows from the above.■

Beyond the indescribable cardinals, we have the measurable cardinals and variants. Particularly, the huge cardinals, the rank-into-rank cardinals, and the supercompact cardinals.

Theorem: If $$\kappa$$ is almost $$n+1$$ huge, then there exists a normal $$\kappa-$$complete almost $$n+1-$$hugness measure $$D$$, closed under $$o_{n-\text{huge}}(X)$$, such that $$\{\lambda\lt\kappa|\lambda\text{ is }n-\text{huge}\}$$. If $$\kappa$$ is $$2^\kappa-$$supercompact, then there exists a normal $$\kappa-$$complete $$2^\kappa-$$supercompactness measure $$D$$, closed under $$o(X)=\{\lambda\lt\kappa|\exists D(D\text{ is a normal measure}\land X\cap\lambda\in D)\}$$.

$$Proof.$$ Let $$j: V\rightarrow M$$ be an almost $$n+1-$$hugeness embedding. Then $$j^{n+1}(\kappa)\gt 2^{j^n(\kappa)}$$, and so the measure generated by $$D$$ is such a measure. Similarly with any $$2^\kappa-$$supercompactness embedding.■

Theorem: If $$I3(\kappa,\lambda)$$, then there exists a normal $$\kappa-$$complete $$I3$$ measure $$D$$ on $$\kappa$$, closed under $$o_{\omega-\text{huge}}(X)$$. If $$I2(\kappa,\lambda)$$, then there exists a normal $$\kappa-$$complete $$I2$$ measure $$D$$ on $$\kappa$$, closed under $$o_I3(X)$$. If $$I1(\kappa,\lambda)$$, then there exists a normal $$\kappa-$$complete $$I1$$ measure $$D$$, closed under $$o_{I2}(X)$$.

$$Proof.$$ Note that the measure witnessing $$n-$$hugeness of $$\kappa$$ is in $$V_\lambda$$, and so $$V_\lambda\vDash\kappa\text{ is }\omega-\text{huge}$$, and the measure generated by $$j: V_\lambda\rightarrow V_\lambda$$ satisfies the above properties.

For the second part.

$$I=\{i: V_\alpha\rightarrow V_\beta\text{ is an elementary embedding with critical point }\kappa|\alpha\lt\beta\land\beta\lt\kappa\}$$.

For $$i,i’\in I$$ with $$i: V_\alpha\rightarrow V_\beta$$ and $$i: V_{\alpha’}\rightarrow V_{\beta’}$$, let $$i\lt^* i’$$ if and only if $$i’\supseteq i$$ and $$\alpha=\beta’$$ and $$\beta’=i(\alpha’)$$.

Let $$\le^*$$ be the reflexive, transitive closure of $$\lt^*$$. Since $$V_\lambda\subseteq M$$, $$I^M=I$$ and $$\le^{*M}=\le^*$$. Any infinite descending sequence in $$M$$ would give as its union the necessary type of embedding. Else, assume it is well-founded in $$M$$. Then it is well-founded in $$V$$. But $$\{j\restriction V_{j^{n+1}(\kappa)}|n\lt\omega\}$$ is an infinite descending sequence. Therefore the measure generated by $$j$$ is such a measure.

For the third part, given an $$I1$$ embedding $$j$$, let $$j’=j\restriction V_\lambda$$. Then $$j’^+(R)=j(R)$$ for an relation $$R$$, and so if $$\lambda_0\lt\lambda$$ and $$I2(\kappa_0,\lambda_0)$$, then $$V_{\lambda+1}\vDash I2(\kappa_0,\lambda_0)$$.■

Theorem: If $$\kappa$$ is supercompact and $$\lambda\gt\kappa$$ is $$n-$$huge, almost $$n-$$huge, $$I3-I1$$, or any $$\Sigma_2$$ property, there exists $$\kappa-$$many $$n-$$ huge, almost $$n-$$huge, $$I3-I1$$, or any $$\Sigma_2$$ property cardinals below $$\kappa$$.

$$Proof.$$ Use the fact that $$\kappa$$ is $$\Sigma_2-$$reflecting, and consider the formula “There exists a $$P$$ cardinal above $$\alpha$$.”■

Let $$o_S=\cap_{\lambda\in Ord} o_\lambda$$, where $$o_\lambda=o_P$$ and $$P$$ is $$\lambda-$$supercompactness.

Theorem: If $$\kappa$$ is extendible, for each $$\lambda$$ there is a $$\lambda-$$extendibility measure $$D$$ on $$\kappa$$ closed under $$o_S(X)$$.

$$Proof.$$ I show for arbitrarily large $$\lambda$$. Let $$\lambda$$, $$\lambda’$$ be inaccessible and let $$j: V_\lambda\rightarrow V_{\lambda’}$$ be a non-trivial elementary embedding with critical point $$\kappa$$. Let $$D$$ be the measure generated by $$j$$, and let $$X\in D$$. Then we need to show that for each $$\alpha\lt\lambda’$$, $$\kappa\in j(o_\alpha(X))$$. Let $$D’$$ be a normal $$\alpha-$$supercompactness measure such that $$X\in D’$$. Then $$D’\in V_{\lambda’}$$, and so $$\kappa\in j(o_\alpha(X))$$.■

2. Additional large cardinals in the large cardinal hierarchy.

There are several interesting identity crisis cases in the lower regions of the large cardinal hierarchy. Namely, the most tragic omission from most lists of large cardinals, the worldly cardinals, can sit above the weakly inaccessible cardinals, by setting $$|\mathbb R|=\kappa$$ for weakly inaccessible $$\kappa$$, but below the inaccessible cardinals. However, if $$\kappa$$ is weakly inaccessible $$L_\kappa\vDash ZFC$$

Theorem: If $$\kappa$$ is inaccessible, there exists a $$\kappa-$$complete normal filter $$F$$ on $$\kappa$$, closed under $$T(X)$$, such that $$\{\lambda\lt\kappa|\lambda\text{ is worldly}\}\in F$$. If $$\kappa$$ is weakly compact, there exists a $$\kappa-$$complete normal filter $$F$$ on $$\kappa$$, closed under $$M(X)$$, such that $$\{\lambda\lt\kappa|\lambda\text{ is Mahlo}\}\in F$$.

$$Proof.$$ A simple alteration to $$^2$$ will suffice. Otherwise, let $$F$$ be the club filter, and let $$C$$ be club. Then there exists a club $$\alpha\lt\kappa$$ such that $$(V_\alpha,\in,C\cap V_\alpha)\vDash C\text{ is club}$$. Finally, to see $$C’=\{\lambda\lt\kappa|\lambda\text{ is worldly }\}$$ is club below $$\kappa$$, take $$C”=\{\lambda\lt\kappa|V_\lambda\prec V_\kappa\}\cap C’$$. Then $$C’\in F$$ and $$C”\supset C’$$. For the second part, take $$\Pi_1^1-$$indescribable filter.■

Right between the inaccessible and Mahlo cardinals in terms of consistency strength, are the pseudo-uplifting and uplifting cardinals.

Theorem: If $$\kappa$$ is pseudo-uplifting, then $$\kappa$$ is a limit of $$\Sigma_3-$$reflecting cardinals, and if $$\kappa$$ is uplifting, then $$\kappa$$ is a limit of pseudo-uplifting cardinals.

$$Proof.$$ See $$^3$$. If $$\kappa$$ is uplifting with target $$\lambda$$, then there must be unboundedly many $$\alpha\gt\kappa$$ such that $$V_\alpha\prec V_\lambda$$, and so $$V_\lambda\vDash(\kappa\text{ is pseudo-uplifting})$$.■

Theorem: If $$\kappa$$ is uplifting and $$\lambda\gt\kappa$$ is strong, superstrong, supercompact, extendible, or any $$\Sigma_3$$ property, there exists $$\kappa-$$many strong, superstrong, supercompact, extendible, or any $$\Sigma_3$$ property cardinals below $$\kappa$$.

$$Proof.$$ Use the fact that $$\kappa$$ is $$\Sigma_3-$$reflecting, and consider the formula “There exists a $$P$$ cardinal above $$\alpha$$.”■

Beyond the hierarchy of indescribable cardinals, the unfoldable are designed to diagonalize over the indescribable cardinals. A cardinal $$\kappa$$ is strongly $$\alpha-$$unfoldable if and only if for every $$S\in H_\kappa$$, there is some $$M\ni S$$ and a non-trivial elementary embedding $$j: M\rightarrow N$$ with critical point $$\kappa$$, $$j(\kappa)\gt\alpha$$, and $$V_\alpha\subseteq N$$.

Theorem: If $$\kappa$$ is $$\kappa+\alpha-$$unfoldable, $$\kappa$$ is $$\alpha-$$indescribable and there is a $$\kappa-$$complete normal filter $$F$$ on $$\kappa$$, closed under $$M(C)$$, such that $$\{\lambda\lt\kappa|\lambda\text{ is }\alpha-\text{indescribable}\}\in F$$.

$$Proof.$$ A simple alteration to $$^4$$ suffices.■

The strongly unfoldable of degree $$\alpha$$ cardinals strengthen the strongly unfoldable cardinals by demanding that $$N\vDash(\kappa\text{ is unfoldable of degree }\lt\alpha)$$. A cardinal is totally unfoldable if and only if it is strongly unfodlable of every degree. Let $$o_\text{unf}(X)=\cap_{\lambda\in Ord} o_{\text{unf}-\alpha}(X)$$, where $$\text{unf}-\alpha$$ is the property of being $$\alpha-$$unfoldable of degree $$\alpha$$. Being unfoldable of every degree can be called total unfoldability.

Theorem: If $$\kappa$$ is strongly unfoldable of degree $$\alpha+1$$, there exists a $$\kappa-$$complete normal filter $$F$$ on $$\kappa$$, closed under $$M(X)$$, such that $$\{\lambda\lt\kappa|\lambda\text{ is totally unfoldable}\}\in F$$. If $$\kappa$$ is superstrongly unfoldable, there exists a $$\kappa-$$complete normal filter $$F$$ on $$\kappa$$, closed under $$o_\text{unf}(X)$$.

$$Proof.$$ Let $$F=\{X\subseteq\kappa|\kappa\in j(X)\}$$. It is immediately normal and $$\kappa-$$complete. Closure follows from $$V_{\kappa+1}\subseteq N$$.

For the second part, we can find some $$N$$ such that $$V_\kappa\prec V_{j(\kappa}$$ and $$V_{j(\kappa}\prec_{\Sigma_3} N$$. Now, let $$F$$ be the filter generated by $$j$$. Assume $$\kappa$$ is strongly unfoldable of degree $$\lt\alpha$$, for $$\alpha\lt j(\kappa)$$. Then the embedding $$j$$ is witnessed by an extender in $$V_{j(\kappa)}$$, and so $$\kappa$$ is $$\lt j(\kappa)-$$unfoldable of degree $$lt j(\kappa)$$, and by reflection is totally unfoldable.

Now, by that argument, for any $$\alpha\lt j(\kappa)$$, if there exists some $$F’\ni X$$ that is an $$\alpha-$$unfoldable of degree $$\alpha$$ filter over $$\kappa$$ as witnessed by an extender, then said extender is in $$V_{j(\kappa)}$$ and so in $$N$$. A reflection argument gives the rest.■

In terms of size, above the measurable cardinals but below the $$I3$$ cardinals are the Woodin and then the superstrong cardinals. Let $$o_\text{sup}(X)=o_P(X)$$, where $$P$$ is superstrongness.

Theorem: If $$\kappa$$ is Woodin, there exists $$\kappa-$$complete normal filter on $$\kappa$$, closed under $$o(X)$$, such that $$\{\lambda\lt\kappa|\lambda\text{ is measurable}\}\in F$$. If $$\kappa$$ is superstrong, then $$\kappa$$ is Woodin and there exists a normal superstrongness measure on $$D$$ on $$\kappa$$, closed under $$o(X)$$, such that $$\{\lambda\lt\kappa|\lambda\text{ is Woodin}\}\in F$$. If $$\kappa$$ is $$1-$$extendible, then $$\kappa$$ is supserstrong, and there exists a normal $$1-$$extendibility measure on $$D$$ on $$\kappa$$, closed under $$o_\text{sup}(X)$$, such that $$\{\lambda\lt\kappa|\lambda\text{ is superstrong}\}\in F$$.

$$Proof.$$ Let $$F$$ be the filter generated by $$C\cap U$$, where $$C$$ is club and $$U=\{\lambda\lt\kappa|\lambda\text{ is measurable}\}$$. $$U$$ is stationary, because if $$C$$ is club and $$f$$ is its corresponding normal function, for some measurable $$\lambda$$, $$f(\lambda)=\lambda$$. We can use a similar argument to get measurable limits of each club. For the rest, a simple alteration to $$^5$$ suffices.■

Let $$o_{\text{E}-\eta}(X)=o_P(X)$$, where $$P$$ is $$\eta-$$extendibility.

Theorem: If $$\eta$$ is not an infinite limit ordinal with $$cf \eta\lt\kappa$$, and $$\kappa$$ is $$\eta+1-$$extendible then $$\kappa$$ is $$\beth_{\kappa+\eta}-$$supercompact and there is a normal $$\eta+2-$$extendiblity measure on $$\kappa$$, closed under $$o_{\beth_{\kappa+\eta}}(X)$$. If $$\kappa$$ is $$\eta+2-$$extendible then $$\kappa$$ is $$\beth_{\kappa+\eta+1}-$$supercompact and there is a normal $$\eta+2-$$extendiblity measure on $$\kappa$$, closed under $$o_{\beth_{\kappa+\eta+1}}(X)$$. If $$\kappa$$ is $$\beth_{\kappa+\eta}-$$supercompact, then there is a normal $$\beth_{\kappa+\eta}-$$supercompactness measure on $$\kappa$$, closed under $$o_{\text{E}-\eta(X)}$$.

$$Proof.$$ For the first part, a simple alteration to $$^6$$ suffices. For the second part, if $$cf\eta\lt\kappa$$ is an infinite limit ordinal, then $$\xi=\eta+1$$ is not a limit ordinal, and so if $$\kappa$$ is $$\xi+1-$$extendible, $$\kappa$$ is $$\beth_{\kappa+\xi}$$supercompact. For the rest, a simple alteration to $$^7$$ suffices.■

It is trivial to see that $$I3$$ cardinals are above this hierarchy, because $$I3(\kappa,\lambda)$$, then $$\kappa$$ is $$\lambda-$$extendible. Furthermore, even Vopěnka cardinals are above this, because if $$\kappa$$ is Vopěnka, then there is a class of $$\lt\kappa-$$extendibles below it.

Theorem: The Vopěnka filter is a normal, $$\kappa-$$complete filter closed under $$o_{\text{E}-\kappa}(X)$$, that contains $$\{\lambda\lt\kappa|V_\kappa\vDash\lambda\text{ is extendible}\}$$. If $$\kappa$$ is almost huge, there exists a normal almost-hugeness measure $$D$$ on $$\kappa-$$ closed under $$o(X)$$, such that $$\{\lambda\lt\kappa|\lambda\text{ is Vopěnka}\}\in D$$. If $$\kappa$$ is huge, there exists a normal hugeness measure $$D$$ on $$\kappa-$$ closed under $$o_\text{almost-huge}(X)$$, such that $$\{\lambda\lt\kappa|\lambda\text{ is almost-huge}\}\in D$$.

$$Proof.$$ A simple alteration to $$^7$$ suffices.■

Theorem: If $$\kappa$$ is strong and $$\lambda\gt\kappa$$ is superstrong or any $$\Sigma_2$$ property, there exists $$\kappa-$$many superstrong or any $$\Sigma_2$$ property cardinals below $$\kappa$$.

$$Proof.$$ Use the fact that $$\kappa$$ is $$\Sigma_2-$$reflecting, and consider the formula “There exists a $$P$$ cardinal above $$\alpha$$.”■

3. Reinhardt cardinals as the ultimate large cardinal

Reinhardt cardinals are, in a sense, the ultimate large cardinal. In consistency strength, they are easily above rank-into-rank cardinals, In size they are weakly extendible and $$\omega-$$huge. However, to my knowledge, they are not necessarily superhuge. The reason for this is that super Reinhardt cardinals are not trivial extensions of Reinhardt cardinals. Let $$o_{S’}(X)=\cap_{\eta\in Ord} o_{\text{E}-\eta}(X)$$.

Theorem: If $$\kappa$$ is superhuge, then there exists a normal hugeness measure on $$\kappa$$, closed under $$o_{S’}(X)$$. If $$\kappa$$ is stationarily superhuge and there exists a Reinhardt or a super Reinhardt cardinal above $$\kappa$$, then there exists $$\kappa-$$many Reinhardt or super Reinhardt cardinals below $$\kappa$$.

$$Proof.$$ Let $$D$$ be the measure generated by a hugeness embedding $$j: V\rightarrow M$$. Then $$M\vDash V_{j(\kappa)}\vDash(\kappa\in j(o_{S’}(X))$$ and so $$\kappa\in j(o_{S’}(X))$$.

For the second part, if $$\kappa$$ is stationarily superhuge, the $$Ord$$ is inaccessible. Therefore the class of all $$\Sigma_1^1-$$reflecting cardinals is club, and so $$\kappa$$ is $$\Sigma_1^1-$$reflecting. Then use the fact that “There exists a Reinhardt cardinal above $$\alpha$$” is $$\Sigma_1^1$$.■

References: 1. Force to change large cardinal strength, Erin Carmody

2. Just how big is the smallest inaccessible cardinal anyway?, Kameryn Williams

3. Resurrection axioms and uplifting cardinals, Thomas A. Johnstone and Joel David Hamkins

4. Chains of End Elementary Extensions of Models of Set Theory, Andres Villaveces

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

6. Extendibility vs supercompactness (Answer), Gabe Goldberg

7. The Higher Infinite, 24. The Strongest Hypothesis, Akihiro Kanamori

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