I have encountered two different definitions of extendible cardinals.

Definition 20.22 of Jech's 2003 Set Theory says:

Definition. A cardinal $\kappa$ is extendible if for every $\alpha > \kappa$ there exist an ordinal $\beta$ and an elementary embedding $j : V_\alpha \to V_\beta$ with critical point $\kappa$.

Meanwhile, Page 311 of Kanamori's 2003 The Higher Infinite says:

Definition. A cardinal $\kappa$ is extendible for all $\eta > \kappa$, there is a $\zeta$ and a $j : V_{\kappa + \eta} \to V_\zeta$ with $\operatorname{crit}(j) = \kappa$ and $j(\kappa) > \eta$ (i.e. $\kappa$ is $\eta$-extendible for all $\eta > \kappa$).

Why are these two definitions equivalent? In particular, it's unclear to me how in Jech's definition, it implies that there exist elementary embeddings with arbitrarily large $j(\kappa)$.


1 Answer 1


Assume ZFC. Suppose $\kappa$ is extendible in Jech's sense, but not Kanamori's. In what follows, all embeddings mentioned will have critical point $\kappa$, without explicitly mentioning it.

Claim 1: There is some $\gamma$ such that for all sufficiently large $\alpha$, if $j:V_\alpha\to V_\beta$ then $j(\kappa)<\gamma$.

Claim 1 is easy to see. Let $\gamma$ be the least witness.

Claim 2: $\gamma$ is a limit ordinal and for all $\xi<\gamma$ and all $\alpha>\kappa$ there is $j:V_\alpha\to V_\beta$ with $j(\kappa)>\xi$.

Proof of Claim 2: Suppose $\gamma=\xi+1$. Then for cofinally many $\alpha\in\mathrm{OR}$, there is $j:V_\alpha\to V_\beta$ with $j(\kappa)=\xi$. It follows that this in fact holds for all $\alpha>\kappa$ (by restricting some embedding with larger domain). Fix an $\alpha>\kappa$ sufficiently large that it witnesses Claim 1, and let $j:V_\alpha\to V_\beta$ be an embedding such that $j(\kappa)=\xi$. Since $\kappa<\alpha\leq\beta$ we can also fix an embedding $k:V_\beta\to V_\delta$ with $k(\kappa)=\xi=j(\kappa)$. Then since $k(\kappa)=\xi<\beta$, we have $k(\xi)>\xi$. But then $\ell=k\circ j:V_\alpha\to V_\delta$ is such that $\ell(\kappa)>\xi$, a contradiction.

Claim 3: If $\beta$ is sufficiently large and $j:V_\beta\to V_\delta$, then $j``\gamma\subseteq\gamma$.

Proof of Claim 3: Suppose not. Then we can fix $\xi<\gamma$ such that for cofinally many $\beta$, hence for all $\beta>\xi$, there is $j:V_\beta\to V_\delta$ such that $j(\xi)\geq\gamma$. Let $\alpha>\xi$ be such that there is some $k:V_\alpha\to V_\beta$ with $k(\kappa)>\xi$, and such that whenever $\ell:V_\alpha\to V_\delta$, then $\ell(\kappa)<\gamma$. Now let $j:V_\beta\to V_\delta$ be such that $j(\xi)\geq\gamma$. Let $\ell=j\circ k:V_\alpha\to V_\delta$. Then $\ell(\kappa)\geq\gamma$, contradiction.

Now let $\alpha$ be sufficiently large and with $\alpha\geq\gamma+2$ and $j:V_\alpha\to V_\beta$. By Claim 3, we have $j``\gamma\subseteq\gamma$. So if $\mathrm{cof}(\gamma)=\omega$ then $j(\gamma)=\gamma$ and $j\upharpoonright V_{\gamma+2}$ contradicts Kunen. But if $\mathrm{cof}(\gamma)>\omega$ then $\lambda=\kappa_\omega(j)$ (the sup of the critical sequence of $j$) is such that ${\lambda<\gamma}$, so $j\upharpoonright V_{\lambda+2}$ contradicts Kunen.

  • $\begingroup$ Thank you. Can you explain why does $\operatorname{cf}(\gamma) = \omega$ imply that $j(\gamma) = \gamma$? $\endgroup$ Mar 15 at 5:57
  • 1
    $\begingroup$ Let $f:\omega\to\gamma$ be cofinal. Then $j(f):j(\omega)\to j(\gamma)$ is cofinal. But $j(\omega)=\omega$. So $j(f)[\omega]$ is cofinal in $j(\gamma)$. But for $n<\omega$, we have $j(f(n))=j(f)(n)$. So $j(f)[\omega]\subseteq\gamma$. So $j(\gamma)=\gamma$. $\endgroup$
    – Farmer S
    Mar 15 at 10:12

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