# Regarding extenders

Let $j\colon M\rightarrow N$ be an elementary embedding (between inner models) with $\operatorname{crit}(j)=\kappa$. Let $\kappa<\lambda$.

Let $\mu$ be the minimal $\alpha$ with $\lambda\le j(\alpha)$ and let $E=\langle E_a\mid a\in[\lambda]^{<\omega}\rangle$ be the $(\kappa,\lambda)$-extender derived from $j$.

Then, for every $\langle a_n\mid n<\omega\rangle$ (with $a_n\in[\lambda]^{<\omega}$) and $\langle X_n\mid n<\omega\rangle$ (with $X_n\in E_{a_n}$), there is a function $f\colon\bigcup\{a_n\mid n<\omega\}\rightarrow\mu$ such that for each $n<\omega$, we have $f"a_n\in X_n$.

I'm interested in a proof/hint for this claim.

Any help would be appreciated, thanks.

• Previously posted at math.stackexchange.com/questions/2740669/regarding-extenders.
– jeq
Apr 23 '18 at 11:37
• As far as I can tell, we need an additional assumption here. May we assume that $E \in M$? Apr 23 '18 at 14:56
• @StefanMesken: It is not assumed. Possibly even for some a's, E_a is not in $M$. Apr 23 '18 at 16:24
• So what you are looking for is a sequence $\langle v_n: n\in \omega\rangle$ such that $v_{n+1}$ projects to $v_n$ the same way $a_{n+1}$ projects to $a_n$ (I'm going to assume $a_n$ here is increasing). The strategy is to show this in $N$ for $\langle j(X_n): n\in \omega\rangle$ and use elementarity to conclude the same in $M$. Even though $\langle a_n: n\in \omega\rangle$ is obviously a witness for $\langle j(X_n): n\in \omega\rangle$ but the sequence may not be in $N$. However, absoluteness of well-foundedness between $V$ and $N$ fixes this (build a tree of attempts ...) Apr 23 '18 at 16:36
• @JingZhang: I'm not assuming that $\langle X_n\mid n<\omega\rangle$ is in M (sounds like you are assuming it (?)). May you expend a bit? Apr 23 '18 at 16:47

Fix $(a_n \mid n < \omega)$, $(x_n \mid n < \omega)$ such that $x_n \in E_{a_n}$ for all $n < \omega$. Without loss of generality we may assume

1. $\{\xi\} \in \{a_n \mid n < \omega \}$ for all $\xi \in \bigcup \{a_n \mid n < \omega\}$ and
2. $a,b \in \{a_n \mid n < \omega\} \implies a \cup b \in \{a_n \mid n < \omega \}$.

(Otherwise close the sequence $(a_n \mid n < \omega)$ under these operations and add dummy values for the corresponding $x$'s.)

Consider the set $T$ of all functions $$t \colon \{a_0, \ldots, a_{n-1} \} \to [\kappa]^{< \omega}$$ such that

1. $\forall i < n \colon t(a_i) \in x_i$ (in particular $\mathrm{card}(t(a_i)) = \mathrm{card}(a_i)$),
2. $\forall i < j < n \colon a_i \cup a_j \in \{ a_0, \ldots, a_{n-1} \} \implies t(a_i \cup a_j) || a_i = t(a_i)$.

The operation $||$ is defined as follows: Let $a \subseteq b$, $B$ be finite sets of ordinals such that $\mathrm{card}(b) = \mathrm{card}(B)$. Write $b = \{b_1 < \ldots < b_k\}$ and $B = \{B_1 < \ldots < B_k \}$. Let $$\pi \colon \mathrm{card}(a) \to b$$ be the unique $<$-preserving function such that $a = \pi " \mathrm{card}(a)$. Then $$B ||a := \{ B_{\pi(0)} < \ldots < B_{\pi(\mathrm{card}(a)-1)} \}.$$

$B || a \subseteq B$ is the subset of $B$ of those elements that correspond to the indexes of $a$'s elements in the increasing enumeration of $b$.

Now consider the tree $(T; \subset)$.

Claim. $(T; \subset)$ is ill-founded.

Proof. Consider $j((T; \subset))$. Since $T$ is countable, elementarity yields that $$j((T; \subset)) = (j " T; \subset)$$ and $$j " T = \{ j(t) \colon \{ j(a_0), \ldots, j(a_{n-1}) \} \to [j(\kappa)]^{< \omega} \mid j(t)(j(a_i)) \in j(x_i) \wedge \ldots \}.$$ In $V$ consider $$b \colon \{ j(a_n) \mid n < \omega \} \to [j(\kappa)]^{< \omega}$$ given by $b(j(a_i)) := a_i$. I'll leave it to you to check that $b \restriction \{j(a_0), \ldots, j(a_{n-1}) \} \in j " T$ for all $n < \omega$ so that $$V \models (j " T; \subset) \text{ is ill-founded}.$$ By absoluteness of wellfoundedness we have that $$N \models j((T; \subset)) = (j"T; \subset) \text{ is ill-founded}$$ and hence, by elementarity, that $$M \models (T; \subset) \text{ is ill-founded}.$$

For the rest of this answer, work in $M$. Let $$b \colon \{a_n \mid n < \omega \} \to [\kappa]^{< \omega}$$ be a branch through $(T; \subset)$. We define $$f \colon \bigcup \{a_n \mid n < \omega \} \to \kappa, \xi \mapsto \bigcup b(\{\xi \}) = \text{ the unique element of } b(\{ \xi \}).$$ (This is possible by our assumption 1. on the sequence $(a_n \mid n < \omega)$.)

Claim. $f$ is as desired.

Proof. Let $a_n = \{\xi_0 < \ldots < \xi_k \}$. Then \begin{align*} f " a_n &= \{ f(\xi_0), \ldots f(\xi_k) \} \\ &= \{ \bigcup b(\{\xi_0\}) , \ldots, \bigcup b (\{\xi_k \}) \} \\ & = \{ \bigcup b(a_n) || \{\xi_0 \}, \ldots, \bigcup b(a_n) || \{\xi_k\} \} \\ &= b(a_n) \in x_n. \end{align*} Q.E.D.

• In this answer I assume that $((a_n,x_n) \mid n < \omega) \in M$ but not that $E \in M$. I also get a slightly better result than required since $f \in M$. If we don't assume that the sequences $(a_n \mid n < \omega), (x_n \mid n < \omega)$ are in $M$, I don't even know where to begin... Apr 23 '18 at 17:56
• Thank you very much for the answer, but i'm more intrested in the general case. Apr 23 '18 at 20:32
• See Kanamori pages 354-355. Apr 23 '18 at 20:42
• @Collapse I agree that what is written is your interpretation but it seems to me that this wasn't intended by the author. As I've suspected, this property is meant to capture well-foundedness of the ultrapower and the version I've proved suffices to conclude that. Maybe the stronger version is true as well, but I remain doubtful about that. Apr 23 '18 at 20:54
• @Collapse: here is a counter example: suppose 0# exists, and $j: L\to L$ with $crit(j)=\aleph_\omega$. (since $V$-cardinals are indiscernibles so this is possible.) Then the $L$-ultrafiler derived from $j$ is not $\omega$-complete (you can translate this to the extender setting), but for each $n$, $\aleph_\omega \backslash \aleph_n+1$ is in the ultrafilter. Apr 23 '18 at 21:08