# Large cardinals: Every weakly compact cardinal has the extension property.

I'm starting to study the basic theory of Large cardinals with Jech's book Set Theory. I'm having struggles understanding the details of the proof that if a cardinal $\kappa$ has the extension property (that is, for every $U\subset V_\kappa$ there exists a transitive set $M$ (with $\kappa\in M$) and $U'\subset M$ such that $(V_\kappa,\in, U)\prec (M,\in ,U')$ ) then it is $\Pi^1_1$-indescribable. That is, given any $\Pi^1_1-$ formula $\forall X\,\varphi(X)$ and $U\subset V_\kappa$ if $\langle V_\kappa, \in, U\rangle \vDash \forall X\,\varphi(X)$ then there exists $\alpha<\kappa$ such that $\langle V_\alpha, \in, U\cap V_\alpha \rangle \vDash \forall X\,\varphi(X)$. For this purpouse Jech takes an arbitrary $\Pi^1_1$ formula and an arbitrary $U\subset V_\kappa$ and making use of the extension property proves that there exists $\alpha<\kappa$ such that $\langle V_\alpha, \in, U\cap V_\alpha \rangle \vDash \forall X\,\varphi(X)$. Here is a screenshot of the proof: Regarding my doubts they are itemized in the following lines:

1. Given $\forall X\varphi(X)$ a $\Pi^1_1$ formula if ($\forall\, X\subset V_\kappa$) $\langle V_\kappa,\in, U\rangle\vDash \varphi(X)$ why $\langle M,\in, U'\rangle \vDash (\forall\, X\subset V_\kappa$) $\langle V_\kappa,\in, U'\cap V_\kappa\rangle\vDash \varphi(X)$? It is because the formula "$\forall\, X\subset V_\kappa$ $\langle V_\kappa,\in, U\rangle\vDash \varphi(X)$" is an absolute? If it is the case, why is absolute?

2. $V^M_\kappa=V_\kappa$: I know that if $\kappa$ is inaccesible then $V_\kappa=L_\kappa$ and since $\kappa\in M$ and $\alpha\mapsto L_\alpha$ is an absolute for transive models then $V^M_\kappa=V_\kappa$. Is it right?

3. If $\langle M,\in, U'\rangle \vDash (\forall\, X\subset V_\kappa$) $\langle V_\kappa,\in, U'\cap V_\kappa\rangle\vDash \varphi(X)$ why $\langle M,\in, U'\rangle \vDash \exists \alpha\; (\forall\, X\subset V_\alpha$) $\langle V_\kappa,\in, U'\cap V_\alpha\rangle\vDash \varphi(X)$

4. The last hence: Why the last line implies $\langle V_\alpha, \in , V_\alpha\cap U\rangle \vDash \sigma$?

Please, Is someone willing to explain me the details under the previous items?

Best,

Cesare

• Why would you think that $V_\kappa = L_\kappa$ for inaccessible $\kappa$? That doesn't hold. – Stefan Mesken Mar 9 '16 at 11:52
• Oh, I hoped so... Thank you for the appointment @Stefan. Then, why the universe $V_\kappa$ is an absolute? – Cesare Mar 9 '16 at 12:06
• I have actually troubles understanding your post. Maybe you could try to phrase your questions more carefully? For example: The model $(M; \in, U')$ depends on the subset $U \subseteq V_\kappa$ that we've chosen. But you seem to fix a single $(M; \in, U')$ throughout your whole post. – Stefan Mesken Mar 9 '16 at 12:17
• Since the rank function is absolute between transitive models $N$ of $\operatorname{ZFC-}$, we have $V_\alpha^N = V_\alpha \cap N$. In our case $V_\kappa \subseteq M$ (since $V_\kappa$ is a substructure) and hence $V_\kappa^M = V_\kappa$. – Stefan Mesken Mar 9 '16 at 21:23
• Looks fine to me. The third one is trivial, but you stated it wrong: $(M; in, U') \models \exists \alpha \forall X \subseteq V_\alpha \colon (V_\alpha; \in, U') \models \phi(X)$ (namely, $\alpha = \kappa$, because $U' \cap V_\kappa = U$ and absoluteness of the modeling relation). Now pull this back to $(V_\kappa; \in, U)$. Then $(V_\kappa; \in , U) \models \exists \alpha \forall X \subseteq V_\alpha \colon (V_\alpha; \in, U)$. By absolutness, $V$ models this as well. – Stefan Mesken Mar 9 '16 at 22:41