In Kanamori's book on large cardinals, he defines a cardinal $\kappa$ to be extendible if for any $\eta$ there is some $j:V_{\kappa+\eta} \prec V_\zeta$ with crit($j$)=$\kappa$ an $j(\kappa)$> $\kappa+\eta$.

On page 323 he is proving 23.15, namely $\kappa$ is extendible iff for any $\eta > \kappa$ there is a $j:V_\eta \prec V_\zeta$ with crit($j$)=$\kappa$. For the proof, assume the latter statement and given $\eta \geq \kappa \cdot\omega$ we wish to prove that $\kappa$ is $\eta$-extendable. He lets $\gamma >\eta$ be such that

(i) if $\beta < \gamma$ and for some $\zeta$ there is $k:V_\eta \prec V_\zeta$ with crit($k$)=$\kappa$ and $k(\kappa$)=$\beta$, there is such a $k$ with a $\zeta < \gamma$.

(ii) cf($\gamma$)=$\omega_1$.

He says that such a $\gamma$ exists by a simple closure argument iterated $\omega_1$ times.

My question is what is the details of this closure argument?

In the paper 'Strong Axioms of Infinity and Elementary Embeddings' by Solovay, Reinhardt and Kanamori, on page 99 they are proving the same thing, and they say that we can get such a $\gamma$ as above by a reflection argument. I would also like to know what is the details of the reflection argument here.


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    $\begingroup$ The definition of extendibility should not say that the critical point is $>\kappa+\eta$ (which makes no sense) but that it equals $\kappa$ and is mapped above $\kappa+\eta$. $\endgroup$ Commented Jun 23, 2022 at 15:44
  • $\begingroup$ That is completely right. Thank you for pointing out the error, I have fixed the definition. Could you share the closure argument and the reflection argument in question? $\endgroup$
    – nana
    Commented Jun 23, 2022 at 17:02

1 Answer 1


The closure argument is actually simple: take $\gamma_0=\eta+1$, $\gamma_\delta=\sup_{\alpha<\delta}\gamma_\alpha$ if $\delta$ is a limit ordinal.

Now define $\gamma_{\alpha+1}$ as follows: For each $\beta < \gamma_\alpha$, choose $k_\beta\colon V_\eta\prec V_{\zeta_\beta}$ such that $\operatorname{crit}k_\beta=\kappa$ and $k_\beta(\kappa)=\beta$ if it exists. Let $X$ be the set of all $\beta<\gamma_\alpha$ such that $k_\beta$ exists, and let $\gamma_{\alpha+1}:=\sup \{\zeta_\beta\mid \beta\in X\}$.

  • $\begingroup$ Thanks! I'm probably missing something simple, why does such a $\nu$ exist? $\endgroup$
    – nana
    Commented Jul 2, 2022 at 14:32
  • $\begingroup$ @kaka Just consider the set of all $\zeta$s (for some $k$) and take the supremum of this set. $\endgroup$
    – Hanul Jeon
    Commented Jul 3, 2022 at 8:30
  • $\begingroup$ Why is it impossible that the $\zeta$'s are unbounded so that there is no supremum? $\endgroup$
    – nana
    Commented Jul 3, 2022 at 15:19
  • $\begingroup$ @kaka Good point and I have to admit that I made a mistake. I will mend my argument. $\endgroup$
    – Hanul Jeon
    Commented Jul 3, 2022 at 17:44
  • $\begingroup$ In your new argument, would you give more details about how the $\gamma_{\alpha+1}$ has the desired feature? If $j:V_\eta \prec V_{\gamma_{\alpha+1}}$, how do we conclude that there is some $j':V_\eta \prec V_\zeta$ with $j(\kappa)=j'(\kappa)$ and $\zeta < \gamma_{\alpha+1}$? My guess is perhaps we want to say that since $j(\kappa) < \gamma_\alpha$, there is some $k_\beta$ that serves as the $j'$? But it may be that $\gamma_\alpha \leq j(\kappa)$, or maybe $\gamma_\alpha > j(\kappa)$, but $\gamma_\alpha$ is just some some $\zeta_\beta$. (Or maybe my guess is misguided.) $\endgroup$
    – nana
    Commented Jul 4, 2022 at 15:33

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