3
$\begingroup$

The following quotes comes from "Cantor's Attic"'s page on supercompacts:

If κ is $|V_{κ+η}|$-supercompact with $η<κ$ then it is preceeded by a stationary set of $η$-extendible cardinals. If $κ$ is $(η+2)$-extendible then it is $|V_{κ+η}|$-supercompact. The least supercompact is not 1-extendible, in fact any cardinal that is both supercompact and 1-extendible is preceeded by a stationary set of cardinals that are both supercompact and limits of supercompact cardinals.

This implies that 1-extendibility is strictly weaker than $2^\kappa$-supercompactness, and weaker than (or as strong as) 3-extendibility. How does 2-extendibility compares to $2^\kappa$-supercompactness? Or more generally $(\eta+1)$-extendibility vs $|V_{\kappa+\eta}|$-supercompactness, $\eta<\kappa$?

Also, is there some known $\eta$ such that $\eta$-extendibility is (consistency-wise) stronger than full supercompactness? By the quote above $\eta\geq\kappa$ is necessary. I'd be surprised to learn that $\eta$-extendibility never gets stronger than supercompactness even though being $\eta$-extendible for all $\eta$ is strictly stronger than supercompactness.

Edit: apparently the quoted results come from Kanamori's "The Higher Infinite", which additionally gives the following interesting result:

If $\kappa$ is $\eta$-extendible and $\delta+5\leq\eta$, then $\kappa$ is $|V_{\kappa+\delta}|$-supercompact.

$\endgroup$
1
  • $\begingroup$ I like how you're one of the main editors of the Cantor's Attic page, and yet you ask about the content here... $\endgroup$
    – Asaf Karagila
    Mar 5, 2018 at 18:45

2 Answers 2

5
$\begingroup$

$2$-extendibility reflects $2^\kappa$-supercompactness. It suffices to show that any $2$-extendible $\kappa$ is $2^\kappa$-supercompact. Then if $\mathcal U$ is a normal fine $\kappa$-complete ultrafilter on $P_\kappa(P(\kappa))$ and $j : V_{\kappa+2}\to V_{\kappa_* + 2}$ witnesses 2-extendibility, $\mathcal U\in V_{\kappa_* + 2}$, and so by the usual reflection argument, $\kappa$ is a limit of $\bar \kappa$ that are $2^{\bar \kappa}$-supercompact.

To demystify Kanamori's result, we show something stronger:

Proposition: If $\kappa$ is $(\eta+1)$-extendible then $\kappa$ is $|V_{\kappa+\eta}|$-supercompact unless $\eta$ is an infinite limit ordinal with $\text{cf}(\eta)< \kappa$.

The case $\eta = 1$ is what we need above. The cofinality restriction is necessary: if $\kappa$ is $|V_{\kappa+\gamma}|$-supercompact where $\gamma$ is a limit ordinal with $\text{cf}(\gamma)< \kappa$, then $\kappa$ is $(\gamma+1)$-extendible but not $|V_{\kappa+\gamma}|$-supercompact in the ultrapower by any $0$-order normal fine $\kappa$-complete ultrafilter on $P_\kappa(|V_{\kappa+\gamma}|)$.

Proof of Proposition. Suppose $j : V_{\kappa+\eta+1} \to V_{\kappa_*+\eta_*+1}$ witnesses $(\eta+1)$-extendibility. We claim that $j$ extends uniquely to an elementary $\hat j : H_{|V_{\kappa+\eta}|^+}\to H_{|V_{\kappa_*+\eta_*}|^+}$. Grant this for now. The cofinality restriction on $\eta$ implies that $|P_\kappa(|V_{\kappa+\eta}|)| = |V_{\kappa+\eta}|$. Therefore $P_\kappa(|V_{\kappa+\eta}|)\in H_{|V_{\kappa+\eta}|^+}$. Hence every subset of $P_\kappa(|V_{\kappa+\eta}|)$ and every function $P_\kappa(|V_{\kappa+\eta}|)\to |V_{\kappa+\eta}|$ are in $H_{|V_{\kappa+\eta}|^+}$. Moreover $\hat j[|V_{\kappa+\eta}|]\in H_{|V_{\kappa_*+\eta_*}|^+}$. So we can set $$\mathcal U = \{X\subseteq P_\kappa(|V_{\kappa+\eta}|) : \hat j[|V_{\kappa+\eta}|]\in \hat j(X)\}$$ and this is a normal fine $\kappa$-complete ultrafilter.

To obtain this extension $\hat j$, use the standard coding fact that for any infinite $\alpha$, there is (uniformly in $\alpha$) a model theoretic interpretation $f_\alpha : V_{\alpha+1}\to H_{|V_\alpha|^+}$. Essentially $f_\alpha(x) = s$ if $x\subseteq V_\alpha$ codes $(E,p)$ where $E$ is a wellfounded extensional relation on a set $A\subseteq V_\alpha$ with $p\in A$, and $\pi(p) = s$ where $\pi : (A,E)\to (M,\in)$ is the transitive collapse of $E$.

Let $f = f_{\kappa+\eta}$ and $f_* = f_{\kappa_*+\eta_*}$. One then sets $\hat j(s) = f_*(j(x))$ for any $x\in f^{-1}(\{s\})$ and checks that $\hat j$ is well-defined and elementary.

Second question. The question is a bit vague: it depends what you mean by "stronger than" and what you mean by "known $\eta$." But anyway the answer is probably no. For example if $\kappa$ is supercompact, then $V_\kappa$ satisfies that for any $\delta$ there is some $\bar \kappa$ that is $\delta$-extendible. Moreover if $\kappa$ is supercompact and $\eta > \kappa$, then there are arbitrarily closed $\kappa$-complete ultrapowers of $V$ in which $\kappa$ is $\eta$-extendible. In particular, if $\varphi(x,y)$ is a $\Sigma_2$ formula with parameters in $V_\kappa$ and there is some $\alpha > \kappa$ such that $\varphi(\kappa,\alpha)$ holds, then $\kappa$ is a limit of cardinals $\bar \kappa$ that are $\bar \alpha$-extendible for some $\bar \alpha < \kappa$ such that $\varphi(\bar \kappa,\bar \alpha)$ holds.

$\endgroup$
2
$\begingroup$

Also, is there some known $\eta$ such that $\eta$-extendibility is (consistency-wise) stronger than full supercompactness?

If $\kappa$ is $\eta$-extendible (or even $\lt\eta$-extendible, that is, $\gamma$-extendible for every $\gamma \lt \eta$) and $\eta$ is the least worldly cardinal greater than $\kappa$ then it follows from Gabe Goldberg's answer that $\kappa$ is $\lt\eta$-supercompact, which implies that $V_\eta \vDash$ ZFC + "$\kappa$ is supercompact" since $\lambda$-supercompactness is downward absolute to $V_{\lambda+2}$. By elementarity of any 1-extendibility embedding, $\kappa$ is a limit of cardinals that are similarly supercompact up to their respective next worldly cardinal.

Added later: For direct implication, suppose $\eta$ is the least $\Sigma_2$-correct cardinal greater than $\kappa$ and $\kappa$ is $\lt\eta$-extendible. Then it is $\lt\eta$-supercompact and by $\Sigma_2$-correctness this is equivalent to full supercompactness.

$\endgroup$
1
  • $\begingroup$ Nice, that does answer the question! In my defense, I think that when I started writing my answer, the phrase "consistency-wise" was not yet part of the question. $\endgroup$ Aug 31, 2021 at 1:39

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.