A question regarding strong cardinals and measure sequence

Question

Let $E$ be a $(\kappa, \lambda)$-extender such that $j: V\to M\simeq Ult(V,E)$ is the corresponding elementary embedding with critical point $\kappa$, $M\supset V_{\kappa+2}$, $M^\kappa\subset M$. Let $\bar{U}$ be a sequence of ultrafilters over $V_\kappa$ of length greater than $\alpha$ where $0<\alpha<(2^\kappa)^+$. Let $E'= E|_{[\alpha]^{<\omega}}$ and $M'\simeq Ult(V,E')$. Notice that $V_{\kappa+2}^\alpha \subset M$ since any transitive closure of such sequence could be coded by a subset of $2^\kappa$ which lies in $V_{\kappa+2}$ so the original set could be recovered by taking transitive collapse in $M$. So $\bar{U}|\alpha \in M$.

My question is:

Is $\bar{U}|\gamma \in M'$ for all $\gamma<\alpha$? I believe it is not true, since if $\alpha=\kappa+1$, then the ultrapower by $E'$ is more or less ultrapower by $U(0)$, by $U(0)\not \in Ult(V,U(0))$ as we know.

The motivation is from a theorem that appeared in the chapter Prikry-type forcings in the Hankbook of Set theory, Lemma 5.1. Where the above is used in the proof to produce a measure sequence of length $\geq ({2^{\kappa} })^+$ from a $\kappa+2$-strong cardinal.

The way I convinced myself is as follows (with GCH): Let $E$ be a $(\kappa, \kappa^{++})$-extender such that $j: V\to M\simeq Ult(V,E)$ is the corresponding elementary embedding with critical point $\kappa$, $M\supset V_{\kappa+2}$, $M^\kappa\subset M$. Just as above $\bar{U}|\alpha \in M$ so there exists $\beta \in [\kappa^{++}]^{<\omega}, f: [\kappa^{++}]^{<\omega}\to V$ such that $j(f)(\beta)=\bar{U}|\alpha$, let $E'= E|_{[\alpha+\sup \beta +1]^{<\omega}}$ then $M'\simeq Ult(V,E')$ will contain $\bar{U}|\alpha$. Since $k: M'\to M$ defined by $k([f]_{U_\gamma})=j(f)(\gamma)$, which ensures that $\bar{U}|\alpha\in ran(k)$. But my concerns are: 1) GCH 2) The length of the restriction of the original extender is very likely to be longer than $\alpha$.

Maybe I'm just missing something basic so any pointer would be appreciated.

Answer 1

Let me explain in a few more details the argument. Let $j_0: V \to V$ and for each $n$, let $j_{n+1}: V \to M_n \simeq M_{n+1}=Ult(M_n, j_n(E))$. Then:

1) $j_{n+1}=j \circ j_n,$

2) $V_{\kappa+2}^\alpha \subset M_n,$ for each $\alpha < (2^\kappa)^+.$

3) $\lambda \geq (2^\kappa)^+.$

Now by induction on $\alpha < (2^\kappa)^+, u_j \restriction \alpha \in M$ ( use clause $(2)$ above).

The main point is to show that for $0<\alpha < (2^\kappa)^+, u_j \restriction \alpha \in \mathcal{U}_\infty.$

Let $\kappa^+ \leq \gamma < (2^\kappa)^+$ and $E' = E \restriction [\gamma]^{<\omega}.$ By clause $(2), E' \in M_n.$ We can factor $j_n$ to get

$j_{E'}: V \to M_{E'} \simeq Ult(V, E')$,

$i_n: M_n \to N_n \simeq Ult(M_n, E')$,

and we will have $i_n = j_{E'} \restriction M_n.$ By elementarity $V_{j_{E'}(\kappa)+2} \cap N_n =V_{j_{E'}(\kappa)+2} \cap M_{E'}. $

Now for $\beta< \gamma, u_j(\beta) =u_{i_n}(\beta)$ so $M_n \models i_n$ constructs $u_j$.

Finally we need to show that $u_j \restriction \alpha \in \mathcal{U_n}.$ This can be proved by induction on $n$, by showing that $u_j \restriction \gamma \in \mathcal{U_n}^{M_k}$ for all $k, \gamma$ with $\gamma+k \leq \alpha.$

Answer 2

I believe the argument could be saved by considering the following ``sub-extender'': For each $\beta<(2^\kappa)^+$, let $a=\bigcup \max\{\beta, \kappa^+\} \cup \bigcup\{a_\gamma: \gamma<\beta\}$ where $a_\gamma \in [\lambda]^{<\omega}$ are those such that there exists $f_\gamma \in V$ such that $\bar{U}\restriction \gamma= j(f_\gamma)(a_\gamma)$ in M. Then we form $Ult(V,E\restriction a)$ to be $M'$ (basic constructions are the same, and we also have coherence property, the final model is formed by taking direct limit as usual). Now we have ensured the relevant objects lie in $M'$. The difference is really that we don't take the entire initial segment to avoid the possibility of blowing up the length of the sub-extender which might make it hard to lie in $M$.