Let us say a cardinal $\kappa$ end-extending if there is a function $F : V_\kappa^{<\omega} \to V_\kappa$ such that: (a) If $M \subseteq V_\kappa$ is closed under $F$, then $M \prec V_\kappa$. (b) If $M$ is closed under $F$ and of size $<\kappa$, then there is $N \supseteq M$ closed under $F$ such that $N \cap \sup(M \cap \kappa) = M \cap \kappa$ and $N \cap \kappa \not= M \cap \kappa$.

It is a standard argument to show that measurable cardinals are end-extending, and this reflects below them. It is fairly easy to see that end-extending cardinals must be regular and cannot be successor cardinals (except for the trivial case $\kappa = \omega_1$).

(1) Is being an end-extending limit cardinal equivalent to a well-known large cardinal notion?

(2) Must end-extending limit cardinals be strongly inaccessible?

  • $\begingroup$ Isn't this something like weakly compact? $\endgroup$
    – Asaf Karagila
    Apr 12, 2019 at 13:00
  • 2
    $\begingroup$ @AsafKaragila Not without collapsing $\omega_1$ it doesn't. $\endgroup$ Apr 12, 2019 at 13:02
  • 3
    $\begingroup$ An end-extending cardinal implies the existence of $0^\sharp$. Just end-extend until you get an elementary submodel of $V_\kappa$ of size $\kappa$. Then the transitive collapse gives you a nontrival $j : L_\kappa \to L_\kappa$. Remarkable and weakly compact cardinals have strength below $0^\sharp$. $\endgroup$ Apr 12, 2019 at 13:11
  • 1
    $\begingroup$ @NotMike I think you can engineer a situation where adding indiscernibles introduces some new small ordinals. If you add some fixed accessible ordinal to each indiscernible, then the resulting set is still indiscernible. But in the argument from a measurable, you add some new ordinal from a certain measure one set. So it’s about adding the “right” indiscernibles. $\endgroup$ Apr 14, 2019 at 11:22
  • 3
    $\begingroup$ I think maybe one can show if $\kappa$ carries a $\omega_1$-saturated ideal $I$ on $\kappa$, then for any $M'\prec V_{\kappa+2}$ containing $I$ of size $<\kappa$, there exists $N\sqsupset M=_{def} M'\cap V_\kappa$ $\endgroup$
    – Jing Zhang
    Apr 16, 2019 at 4:58

1 Answer 1


Suppose $\kappa$ carries an $\omega_1$-saturated $\kappa$-complete ideal $I$, given $M\prec (V_{\kappa+2},\in , <)$ ($<$ well orders $V_{\kappa+2}$) of size $<\kappa$ containing $I$, we show how to find an end-extension $N\prec V_{\kappa+2}$ of $M$ (i.e. $N\cap \sup(M\cap \kappa)=M\cap \kappa$).

Let $G\subset P(\kappa)/I$, and let $j: V\to N$ be the generic embedding in $V[G]$. Look at $M'=\{j(f)(\kappa): f\in M\}$. In $N$, $M'\prec j((V_{\kappa+2}, \in , <))$. Clearly $\kappa\in M'\cap j(\kappa)- j(M)$. Note $j(M)\cap j(\kappa)=j''M \cap j(\kappa)=j(M\cap \kappa)=j'' M\cap \kappa=M\cap \kappa$. If $\xi\in \kappa$ such that $\xi\in M'$, $\xi\in M$. The reason is that: fix $f$ such that $j(f)(\kappa)=\xi$. In $V$, $M$ contains a maximal antichain $\{A_i: i<\omega\}$ such that for each $i$, there exists $\delta_i<\kappa$, $A_i\Vdash j(f)(\kappa)=\delta_i$. Clearly $\{A_i: i<\omega\}, \{\delta_i: i<\omega\}\subset M$. Hence in $V[G]$, there exists $i<\omega$, $\xi=\delta_i \in M$. This shows $M'\cap \sup (j(M)\cap j(\kappa))=M'\cap \sup(M\cap \kappa)=M\cap \kappa=j(M)\cap j(\kappa)$.

By elementarity, there exists $N\prec V_{\kappa+2}$ containing $M$ such that $N\cap \sup (M\cap \kappa)=M\cap \kappa$ and $N\cap \kappa \neq M\cap \kappa$.

So $\kappa$ can be $2^\omega$.

  • 1
    $\begingroup$ This also works to show that for $\mu$-c.c. ideals, $\mu<\kappa$, almost all $M$ such that $M \cap \mu \in \mu$ can be end-extended. $\endgroup$ Apr 16, 2019 at 20:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.