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# $$A_6^*$$ is equivalent to $$A_6$$ [WIP]

In there paper Strong Axioms of Infinity and Elementary Embedding, Robert Solovay, William Reinhardt, and Akihiro Kanamori introduced the axioms $$A_1$$-$$A_7$$, measuring the gap between hugeness and extendibility.

$$A_1(\kappa)$$: $$\kappa$$ is huge.

$$A_2(\kappa)$$: There is some $$j: V_\alpha\prec V_\beta$$ with critical point $$\kappa$$ such that $$j(\kappa)\le\alpha$$.

$$A_3(\kappa)$$: $$\kappa$$ is almost huge; i.e. there is some $$j: V\prec M$$ with critical point $$\kappa$$ such that $$M^\lambda\subseteq M$$ for every $$\lambda\lt j(\kappa)$$; we call this $$M^{\lt j(\kappa)}\subseteq M$$.

$$A_4(\kappa)$$: There is some $$\lambda\gt\kappa$$ and a normal ultrafilter $$U$$ over $$P_\kappa(\lambda)$$ such that if $$f: \kappa\rightarrow\kappa$$, then $$j_U(f)(\kappa)\lt\lambda$$.

$$A_5(\kappa)$$: There is a normal ultrafilter $$U$$ over $$\kappa$$ such that for any natural sequence of structure $$\langle\mathfrak M_\zeta|\zeta\lt\kappa\rangle$$ each in $$V_\kappa$$, there is some $$X\in U$$ such that if $$\alpha\lt\beta\in X$$ there is some $$j: \mathfrak M_\alpha\prec\mathfrak M_\beta$$ such that $$\text{crit}j=\alpha$$.

$$A_6(\kappa)$$: $$V_{\kappa+1}$$ is a natural model of $$KM$$ set theory $$+$$Vopěnka’s principle.

$$A_6^*(\kappa)$$: $$V_{\kappa+1}$$ is a natural model of $$KM$$ set theory and there is a class $$S$$, stationary in $$\kappa$$, such that if $$\alpha_0\lt…\lt\alpha_n\lt\beta_0\lt…\lt\beta_n\in S$$, then there is some $$j: V_{\alpha_n}\prec V_{\beta_n}$$ such that $$j(\alpha_i)=\beta_i$$.

$$A_7(\kappa)$$: $$\kappa$$ is extendible and there is a normal ultrafilter over $$\kappa$$ such that $$\{\alpha\lt\kappa|\alpha\text{ is extendible}\}\in U$$.

If $$A_1(\kappa)$$, there is a normal ultrafilter $$U$$ over $$\kappa$$ such that $$\{\alpha\lt\kappa|A_2(\alpha)\}\in U$$; if $$A_2(\kappa)$$, then $$A_3(\kappa)$$ and there is a normal ultrafilter $$U$$ over $$\kappa$$ such that $$\{\alpha\lt\kappa|A_3(\alpha)\}\in U$$; and if $$A_3(\kappa)$$, then $$A_4(\kappa)$$ and there is a normal ultrafilter $$U$$ over $$\kappa$$ such that $$\{\alpha\lt\kappa|A_4(\alpha)\}\in U$$.

The first part is simple; take some $$j: V\prec M$$ witnessing the hugeness of $$\kappa$$; then $$|j\restriction V_{j(\kappa)}|=|V_{j(\kappa)}|=j(\kappa)$$ and so $$j\restriction V_{j(\kappa)}: V_{j(\kappa)}\prec V_{j^2(\kappa)}^M$$ witnesses $$A_2(\kappa)$$ in $$M$$, and so $$\{\alpha\lt\kappa|A_2(\alpha)\}\in\{X\subseteq\kappa|\kappa\in j(X)\}$$.

The second and third parts are trickier, but with manipulation of sequences of ultrafilters and ultrafilters respectively, we can get the result we want.

We call a cardinal $$\kappa$$ Vopěnka if $$A_6(\kappa)$$. If $$\kappa$$ is Vopěnka and $$U$$ is a normal ultrafilter over $$\kappa$$, then $$\{\alpha\lt\kappa|\alpha\text{ is Vopěnka}\}\in U$$, because $$V_{\kappa+1}\subseteq \text{Ult}_U(V)$$ and the Vopěnkaness of $$\kappa$$ is a property of $$V_{\kappa+1}$$. In particular, if $$U$$ witnesses $$A_5(\kappa)$$, then $$\{\alpha\lt\kappa|\alpha\text{ is Vopěnka}\}\in U$$.

Furthermore, if $$\alpha\in S$$ and $$S$$ witnesses $$A_6^*(\kappa)$$, then $$V_\alpha\prec V_\kappa$$ and $$A_7(\alpha)$$; i.e. $$\alpha$$ is extendible and there is a normal ultrafilter over $$\kappa$$ such that $$\{\beta\lt\alpha|\beta\text{ is extendible}\}\in U$$.

Proposition: $$A_6^*(\kappa)$$ if and only if $$\kappa$$ is Vopěnka.

$$Proof.$$ For the forward direction, we use the equivalent definition of Vopěnkaness; $$\kappa$$ is Vopěnka if and only if for every $$A\subseteq V_\kappa$$, there is some $$\alpha$$ such that for every $$\eta\lt\kappa$$, there is some $$j: (V_{\kappa+\eta},\in,A\cap V_{\kappa+\eta})\prec (V_\zeta,\in,A\cap V_\zeta)$$. Fix some $$A\subseteq V_\kappa$$. Then $$C=\{\alpha\lt\kappa|(V_\alpha,\in,A\cap V_\alpha)\prec(V_\kappa,\in,A\cap V_\kappa)\}$$ is club in $$\kappa$$, because $$\kappa$$ is inaccessible, and so fix any $$\alpha_0\in C\cap S$$ and a sequence $$\alpha_0\lt\alpha_0+\eta\lt\alpha_1…\lt\alpha_n\lt\beta_0\lt…\lt\beta_n\in C\cap S$$.

Then $$(V_\kappa,\in,A)\vDash\phi(x)$$ if and only if $$(V_{\alpha_n},\in,V_{\alpha_n}\cap A)\vDash\phi(x)$$ if and only if $$(V_{\beta_n},\in,A_0)\vDash\phi(j(x))$$ if and only if $$(V_{\beta_n},\in,A\cap V_{\beta_n})\vDash\phi(j(x))$$. Then $$j\restriction V_{\alpha+\eta}: (V_{\kappa+\eta},\in,A\cap V_{\kappa+\eta})\prec (V_\zeta,\in,A\cap V_\zeta)$$.

For the converse direction, let $$F_{\text{Vop},\kappa}$$ be the Vopěnka filter, and $$S_\sigma$$ be defined by induction along the length of $$\sigma$$, for $$\sigma\in\kappa^{\lt\omega}$$. If $$|\sigma|=n+1$$, $$n$$ is even and $$\sigma(n)\in S_{\sigma\restriction n}$$, then let:

$$S_\sigma=\{\alpha\lt\kappa|\exists j: V_{\sigma(n/2)}\prec V_{\sigma(n/2)}(\text{crit}j=\alpha_0\land\forall i\lt n(j(\sigma(i))=\sigma(i+n/2))\}$$

If $$n$$ is odd:

$$S_\sigma=\{\alpha\lt\kappa|\exists j: V_{\sigma((n+1)/2)}\prec V_{\sigma((n+1)/2)}(\text{crit}j=\alpha_0\land\forall i\lt n(j(\sigma(i))=\sigma(i+(n+1)/2))\}$$

Else $$S_\sigma=\kappa$$. We prove $$S_\sigma\in F_{\text{Vop},\kappa}$$.

Norman Perlmutter introduced a series of large cardinals designed to measure the gap between almost hugeness and supercompactness. If $$j: V\prec M$$ is an elementary embedding, then let the clear of $$j$$ $$\theta=\text{sup}\{j(f)(\text{crit}j)|f: \text{crit}j\rightarrow \text{crit}j\}$$. $$\kappa$$ is high-jump if and only if there is some $$j: V\prec M$$ such that $$M^\theta\subseteq M$$ for $$\theta$$ the clearance of $$j$$ and $$\kappa$$ the critical point $$j$$. Respectively, $$\kappa$$ is Shelah for supercompact if and only if for every function $$f: \kappa\rightarrow\kappa$$, there is some $$j: V\prec M$$ such that $$M^{j(f)(\kappa)}\subseteq M$$ and $$\text{crit}j=\kappa$$.

Proposition: If $$\kappa$$ is Shelah for supercompactnes, then $$A_6(\kappa)$$, and there is a normal ultrafilter $$U$$ such that $$\{\alpha\lt\kappa|A_6(\kappa)\}\in U$$.

$$Proof.$$ By a result of Permultter, $$A_6(\kappa)$$ if and only if $$\kappa$$ is Woodin for supercompactness, and because $$\kappa$$ is Shelah for supercompactness, $$\kappa$$ is Woodin for supercompactness. Also, given any $$U$$ a normal ultrafilter over $$\kappa$$, $$V_{\kappa+1}\subseteq\text{Ult}_U(V)$$ and so $$\text{Ult}_U(V)\vDash A_6(\kappa)$$, and therefore $$\{\alpha\lt\kappa|A_6(\kappa)\}\in U$$.■

Proposition: $$\kappa$$ is $$2-$$fold Shelah if and only if $$\kappa$$ is Shelah for supercompactness.

Proposition: If $$A_5(\kappa)$$, as witnessed by $$U$$, then $$\{\alpha\lt\kappa|\alpha\text{ is Shelah for supercompactness}\}\in U$$.

Proposition: $$A_4(\kappa)$$ if and only if $$\kappa$$ is high-jump.

What about the lower reaches of Permultter’s hierarchy?

Proposition: $$\kappa$$ is enhanced supercompact if and only if it is extendible and there is a strong cardinal above it.

Corollary: If $$A_7(\kappa)$$, as witnessed by $$U$$, then $$\{\alpha\lt\kappa|\alpha\text{ is enhanced supercompact}\}\in U$$.

Proposition: $$\kappa$$ is $$(2^\kappa)^+-$$hypercompact if and only if it is hypercompact.

Proposition: If $$\kappa$$ is extendible, it is hypercompact.

Corollary: If $$\kappa$$ is enhanced supercompact, it is hypercompact.

Proposition: If $$\kappa$$ is $$\beta+1-$$hypercompact and $$\beta\lt (2^\kappa)^+$$, then $$\kappa$$ is excessively $$\beta-$$hypercompact and there is a normal ultrafilter $$U$$ such that $$\{\beta\lt\kappa|\beta\text{ is excessively }\beta-\text{hypercompact}\}\in U$$.

Proposition: If $$\kappa$$ is excessively $$\beta-$$hypercompact and $$\kappa^+\lt\beta\lt (2^\kappa)^+$$, then there is a normal ultrafilter $$U$$ such that $$\{\beta\lt\kappa|\beta\text{ is }\beta-\text{hypercompact}\}\in U$$.

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