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In Kanamori's Book The Higher Infinite is proved in Theorem 7.8. (implication (b) $\rightarrow$ (a) ) that every inaccesible cardinal with the tree property is a weakly compact cardinal by means of the extension property.

The idea of the proof is bild up a $\kappa-$tree (which will have a branch of size $\kappa$), consider the direct limit of one of its branches and finally take the transitive collapse of that limit of models. I'm having troubles understanding how Kanamori is able to ensure that the cardinal $\kappa$ lies in that transitive collpase $\langle X,\in, R\rangle $. He says that every $\beta_\xi$ collapse to the same ordinal $\beta\in X$ and thus this $\beta$ must be greater or equal than $\kappa$. I adjoint a screenshot of the proof:

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I don't see why every ordinal $\beta_\xi$ collapses to the same ordinal and moreover how it implies that $\beta\geq \kappa$. Could someone explain me the details of this last part?

Best,

Cesare.

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  • $\begingroup$ Do you know how to form the direct limit? This is an immediate consequence of its construction. $\endgroup$ – Stefan Mesken Mar 11 '16 at 12:33
  • $\begingroup$ I realized that my previous comment may send the wrong message. What I meant to say is: In case you know how to construct the direct limit, but don't see how these facts follow, I'm happy to provide an answer. In case you don't know how to construct a direct limit, you should consult a textbook. $\endgroup$ – Stefan Mesken Mar 11 '16 at 12:43
  • $\begingroup$ @Stefan The direct limit is formed as $\bigcup_{\xi<\kappa} i_\xi(H(\xi,\beta_\xi))$ where $i_\xi$ is an elementary embedding between $H(\xi,\beta_\xi)$ and the direct limit. Is it right? If it is, could you explain me the details of my question? $\endgroup$ – Cesare Mar 11 '16 at 13:36
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Let $( [H(\xi, \beta_\xi)] \mid \xi < \beta)$ be a cofinal branch in $T$ and let $M$ be the direct limit of the linear system $( H(\xi, \beta_\xi), i_{\xi, \eta} \mid \xi \le \eta < \kappa)$, where $i_{\xi, \eta} \colon H(\xi, \beta_\xi ) \to H(\eta, \beta_\eta)$ is an elementary embedding such that $i_{\xi,\eta} \restriction V_{\alpha_\xi} = \operatorname{id}$ and $i_{\xi, \eta}(\beta_\xi) = \beta_\eta$.

We identify $M$ with its transitive collapse. For each $\xi < \kappa$ let $\pi_\xi \colon H(\xi, \beta_\xi) \to M$ be the elementary embedding into the direct limit.

We first claim that there is an ordinal $\beta$ such that $\pi_\xi(\beta_\xi) = \beta$ for all $\xi < \kappa$:

Proof. Since $H(\xi, \beta_\xi) \models \beta_\xi \text{ is an ordinal}$ and $\pi_\xi$ is elementary, we have that $M \models \pi_\xi(\beta_\xi) \text{ is an ordinal}$. Now $M$ is transitive and 'being an ordinal' is a $\Sigma_0$-property. Thus $\beta := \pi_\xi(\beta_\xi)$ is an actual ordinal.

We claim that $\beta$ is independent of $\xi < \kappa$: Let $\xi \le \eta < \kappa$. Since $M$ is a direct limit we have $\beta = \pi_\xi(\beta_\xi) = \pi_\eta \circ i_{\xi, \eta}(\beta_xi) = \pi_\eta( \beta_\eta)$. Q.E.D.

Next, we claim that $\beta \ge \kappa$:

Proof. Assume, toward a contradiction, that $\beta < \kappa$. Since $( [H(\xi, \beta_\xi)] \mid \xi < \beta)$ is cofinal in $T$, there is some $\xi < \kappa$ such that $\beta < \alpha_\xi$. Since $H(\xi, \beta_\xi) \cap V_{\alpha_\xi}$ is transitive, we have $\beta \in H(\xi, \beta_\xi)$. By the elementarity of $\pi_\xi$ we now have $\pi_\xi(\beta) < \pi_\xi(\beta_\xi) = \beta$. This is impossible, because for any homomorphism $$ \rho \colon (N;\in) \to (N'; \in) $$ between transitive structures and any ordinal $\gamma \in N$, we always have $\rho(\gamma) \ge \gamma$. (Otherwise let $\delta$ be the least ordinal such that $\rho(\delta) < \delta$. Then $\rho(\delta) \in N$ and, by because $\rho$ is a $\in$-homomorphism, $\rho(\rho(\delta)) < \rho(\delta) < \delta$. Thus $\delta' := \rho(\delta)$ satisfies $\rho(\delta') < \delta'$ - contradicting the choice of $\delta$.) Q.E.D.

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  • $\begingroup$ I had to modify my proof slightly, because the $H(\xi, \beta_\xi)$ aren't transitive. Since they contain a sufficiently large transitive structure, our argument still goes through (if I guessed correctly that the $\alpha_\xi$ are cofinal in $\kappa$). $\endgroup$ – Stefan Mesken Mar 11 '16 at 14:12
  • $\begingroup$ Thank you very much @Stefan ! Regarding to $H(\xi,\beta_\xi)$ they aren't transitive but $V_{\alpha_\xi}$ does and $V_{\alpha_\xi}\subset H(\xi,\beta_\xi)$ by definition. So I think your proof still works :) $\endgroup$ – Cesare Mar 11 '16 at 14:21
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    $\begingroup$ @Cesare That's what my comment is about ^^ $\endgroup$ – Stefan Mesken Mar 11 '16 at 14:22

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