I asked this at math.stackexchange, but got no answers:

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.

  • $\begingroup$ Previously posted at math.stackexchange.com/questions/2740669/regarding-extenders. $\endgroup$
    – jeq
    Apr 23 '18 at 11:37
  • $\begingroup$ As far as I can tell, we need an additional assumption here. May we assume that $E \in M$? $\endgroup$ Apr 23 '18 at 14:56
  • $\begingroup$ @StefanMesken: It is not assumed. Possibly even for some a's, E_a is not in $M$. $\endgroup$
    – Collapse
    Apr 23 '18 at 16:24
  • 1
    $\begingroup$ 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 ...) $\endgroup$
    – Jing Zhang
    Apr 23 '18 at 16:36
  • $\begingroup$ @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? $\endgroup$
    – Collapse
    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.

  • $\begingroup$ 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... $\endgroup$ Apr 23 '18 at 17:56
  • $\begingroup$ Thank you very much for the answer, but i'm more intrested in the general case. $\endgroup$
    – Collapse
    Apr 23 '18 at 20:32
  • 1
    $\begingroup$ See Kanamori pages 354-355. $\endgroup$
    – Collapse
    Apr 23 '18 at 20:42
  • 1
    $\begingroup$ @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. $\endgroup$ Apr 23 '18 at 20:54
  • 5
    $\begingroup$ @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. $\endgroup$
    – Jing Zhang
    Apr 23 '18 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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