From $\textit{The Higher Infinite}$ by Kanamori, in his proof (page 39 - 41) that $\kappa$ is weakly compact if and only if $\kappa$ satisfies the Extension Property, the proof does something that I am not entirely sure is necessary:

The relevant portion of this proof is as follows:

Let $\Sigma$ be a $\kappa$ satisfiable set of $L_{\kappa,\kappa}$ sentences. $R_1$ is the definable satisfaction relation for set models of $V_\kappa$. $R_2$ is another relation (but that is not relevant at this point in the proof). He then manages to prove that there is a transitive set $X$ properly containing $V_\kappa$ such that

$\langle V_\kappa, \in, R_1, R_2 \rangle \prec \langle X, \in, S_1, S_2 \rangle$

$\langle X, \in, S_1, S_2) \models \Sigma \text{ has a model }$

I think that the proof should be finished at this point: Let $\mathcal{M}$ in $X$ be such a model. By the absoluteness of the satisfaction relation $S_1$, $\mathcal{M}$ is a model of $\Sigma$ in $V$. This should conclude the proof.

However, Kanamori continues. Since $\kappa$ is inaccessible, $V_\kappa \models ZFC$. By elementarily, $X \models ZFC$. Being inaccessible is absolute for transitive model of $ZFC$, so $X \models \kappa$ is inaccessible. Then he applies the Lowenheim Skolem Theorem in $X$ for $L_{\lambda, \lambda}$ for inaccessible cardinal $\lambda$ to prove that (when $\lambda = \kappa$)

$\langle X, \in, S_1, S_2 \rangle \models \Sigma$ has a model $\mathcal{M}$ with domain $\subseteq \kappa$

Then he applies the absoluteness of the satisfaction relation $S_1$, to claim that $\mathcal{M}$ is really a model of $\Sigma$.

My question is why does he have to produce a model in $X$ of $\Sigma$ with domain a subset of $\kappa$ to apply the absoluteness argument of the satisfaction relation. Would it not be enough just to produce any model in $X$. Thanks for any clarification

  • $\begingroup$ There are about a zillion ways to define weakly compact cardinals, which one are you using? And also is the extension property the existence of elementary end-extensions? $\endgroup$ – Asaf Karagila Mar 30 '13 at 7:12
  • $\begingroup$ Yes. The extension property for $\kappa$ is the existence of a transitive $X$ a proper subset of $V_\kappa$ and $S \subset X$ such that $(V, \in, R) \prec (X, \in, S)$. The definition of weakly compact is the $\kappa$ "compactness theorem" for collections of sentences using no more than $\kappa$ no logical sentences. $\endgroup$ – William Mar 30 '13 at 7:30
  • $\begingroup$ It's a shame on the bounty. Perhaps it'd be time to cross this to MathOverflow when the bounty expires? $\endgroup$ – Asaf Karagila Apr 12 '13 at 0:31

The problem is that you cannot conclude "$\mathcal{M}$ is a model of $\Sigma$" from "$X \vDash \mathcal{M}$ is a model of $\Sigma$". The reason is that there is no reason why satisfaction of an $L_{\kappa, \kappa}$ sentence in $\mathcal{M}$ as seen by $X$ will be truly reflected in $V$ unless $X$ can see all of $\mathcal{M}^{< \kappa}$. Choosing the universe of $\mathcal{M}$ to be a subset of $V_{\kappa}$, achieves this.

This issue does not arise in the usual $L_{\omega, \omega}$-satisfaction since every decent transitive class is closed under finite sequences.


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