# Weakly Compact Cardinals are Mahlo Proof

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

• 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 :) – BrianO Apr 5 '16 at 23:43
• Thanks Brian, that makes sense. Probably that $F$ is unbounded in $C$? – Jmaff Apr 5 '16 at 23:53
• 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. – BrianO Apr 6 '16 at 0:03
• Ah, thank you Brian. I understand and I added the cofinality assumption you mentioned. – Jmaff Apr 6 '16 at 0:10

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 $\lambda$.
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.