Timeline for Showing that $\alpha$ isn't a cardinal in $J_{\alpha+1}^{\vec E}$ for a fine extender sequence $\vec E$

6 events

when toggle format	what		by	license	comment
Jun 22, 2016 at 20:55	comment	added	Stefan Hoffelner		Yes you are right.
Jun 22, 2016 at 20:43	vote	accept	Dan Saattrup Nielsen
Jun 22, 2016 at 20:42	comment	added	Dan Saattrup Nielsen		Awesome. I have one last question, if you don't mind. To say that F is definable with parameters from $J_\alpha^{\vec E}$ requires that $(E_\alpha)_a\in J_\alpha^{\vec E}$ for any a, no? Are we using the amenable encoding of $E_\alpha$ to achieve this?
Jun 22, 2016 at 19:59	comment	added	Stefan Hoffelner		Yes your $x \in P(\kappa)$ codes the function which in turn, together with $a \in [\nu]^{<\omega}$ represents an ordinal less than $\alpha$ in the ultrapower. (Using that all the information coded in $E_{\alpha}$ is already in the extender restricted to its support.)
Jun 22, 2016 at 15:57	comment	added	Dan Saattrup Nielsen		Thanks for the answer Stefan! Am I understanding you correctly if the surjection, let's call it F, is then defined as $F(x,a)=\beta$ iff $Ult\vDash [f_x,a]=\beta$ where $f_x:[\kappa]^{\|a\|}\to\kappa$ corresponds to $x\in P(\kappa)$ and $\beta<\alpha$? Equivalently $F(x,a)=\beta$ iff $f_x^{a,a\cup\{\beta\}}(u)=\bigcup^{\{\beta\},a\cup\{\beta\}}(u)$ for $(E_\alpha)_{a\cup\{\beta\}}$ a.e. $u$?
Jun 22, 2016 at 15:31	history	answered	Stefan Hoffelner	CC BY-SA 3.0