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?



End-extending cardinals
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 $2^\omega$.


