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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:

enter image description here

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

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  • $\begingroup$ Why would you think that $V_\kappa = L_\kappa$ for inaccessible $\kappa$? That doesn't hold. $\endgroup$ – Stefan Mesken Mar 9 '16 at 11:52
  • $\begingroup$ Oh, I hoped so... Thank you for the appointment @Stefan. Then, why the universe $V_\kappa$ is an absolute? $\endgroup$ – Cesare Mar 9 '16 at 12:06
  • $\begingroup$ 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. $\endgroup$ – Stefan Mesken Mar 9 '16 at 12:17
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    $\begingroup$ 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 $V_\kappa^M = V_\kappa$. $\endgroup$ – Stefan Mesken Mar 9 '16 at 21:23
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    $\begingroup$ 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. $\endgroup$ – Stefan Mesken Mar 9 '16 at 22:41

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