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I have a question about a corollary in Jech's set theory text which states:

Corollary 17.19. Every Weakly Compact cardinal $ \kappa $ is a Mahlo cardinal, and the set of Mahlo cardinals below $ \kappa $ is stationary.

I have a question about the proof of the first part . In particular,

Proof : Let $ C \subset \kappa $ be a closed unbounded set. Since $ \kappa $ is inaccessible , $ ( V_{ \kappa} , \in , C)$ satisfies the $ \Pi_{1}^{1} $ sentence:$ \not\exists F ( F \text{ is a function from some } \lambda < \kappa \text{ cofinally into } \kappa )$ and $C$ is unbounded in $ \kappa$.

I am confused as to why this last line is true. In particular , doesn't $ \kappa $ having rank $ \kappa $ imply that $V_{\kappa}$ cannot even interpret the sentence? Am I misunderstanding what is meant by this line in the first place? Thanks

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    $\begingroup$ True, there's no need (and, in $V_{\kappa}$, no way) to mention $\kappa$. The sentence should just say "$\neg\exists F\,(\text{$F$ is a function, $dom(F)$ is an ordinal, and $C$ is unbounded)}$". I'll have to look at the proof — it seems the sentence ought to say that $C$ and $F$ have something to do with each other :) $\endgroup$
    – BrianO
    Commented Apr 5, 2016 at 23:43
  • $\begingroup$ Thanks Brian, that makes sense. Probably that $F$ is unbounded in $C$? $\endgroup$
    – Jmaff
    Commented Apr 5, 2016 at 23:53
  • $\begingroup$ I looked: to be less confusing, the sentence should say "$\neg\exists F\,(\text{$F$ is a function, $dom(F)$ is an ordinal, and $range(F)$ is unbounded), and $C$ is unbounded}$". (You left out the "cofinally" part: Jech writes "... $F$ maps $\lambda < \kappa$ cofinally into $\kappa$...".) In fact no connection between $F$ and $C$ is needed. By $\Pi_1^1$ indescribability, there is a regular $\alpha<\kappa$ such that $(V_{\alpha}, \in, C\cap V_{\alpha}) \models$ that sentence, so $C\cap V_{\alpha}$ is unbounded (in $\alpha$), thus $\alpha = sup C\cap V_{\alpha} \in C$ as C ls club. $\endgroup$
    – BrianO
    Commented Apr 6, 2016 at 0:03
  • $\begingroup$ Ah, thank you Brian. I understand and I added the cofinality assumption you mentioned. $\endgroup$
    – Jmaff
    Commented Apr 6, 2016 at 0:10

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In $V_\kappa$ it is very easy to understand what is $\kappa$. It's the class of all the ordinals. So the sentence is really "For every relation over $V_\kappa$ which is a function with domain being an ordinal, the range is bounded".

  1. For every relation over $V_\kappa$, since if $\kappa$ were singular, then a function witnessing that is not an element of $V_\kappa$, but rather a subset of $V_\kappa$.

  2. The domain is an ordinal simply means that for some $\lambda<\kappa$, the domain is this $\lambda$.

  3. The range is bounded means that the function is not cofinal.

Alternatively, you could say that for every subset of $V_\kappa$ which is a function with domain being an element of $V_\kappa$, there is an element of $V_\kappa$ which realizes this function entirely. I'll leave you to think about why this alternative sentence works too.

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