3
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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 Apr 5 '16 at 23:43
  • $\begingroup$ Thanks Brian, that makes sense. Probably that $F$ is unbounded in $C$? $\endgroup$ – Jmaff Apr 5 '16 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 Apr 6 '16 at 0:03
  • $\begingroup$ Ah, thank you Brian. I understand and I added the cofinality assumption you mentioned. $\endgroup$ – Jmaff Apr 6 '16 at 0:10
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.