I have been reading Kanamori's Higher Infinite and I am trying to understand that a cardinal $\kappa$ is $\Pi^1_1$-indescribable iff it has the extension property.

We say that $\kappa$ has the extension property iff for every $R \subseteq V_\kappa$ there are sets $X$, $S$ such that $X \neq V_\kappa$ and $S \subseteq X$, such that $\langle V_\kappa, \in, R \rangle \prec \langle X, \in, S\rangle$.

This sounds like a rather easy property to satisfy, can someone help me acquire some intuitive insight why this is not satisfied for $\kappa$s smaller than the $\Pi^1_1$-indescribabile cardinal?

EDIT: A cardinal $\kappa$ is $\Pi^1_1$-indescribable iff for every $\Pi^1_1$ formula $\varphi$ and a set $R \subseteq V_\kappa$ with $\langle V_{\kappa+1}, \in, R \rangle \models \varphi$, there is an ordinal $\alpha$ such that $\langle V_{\alpha+1}, \in, R \cap V_\alpha \rangle \models \varphi$.

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    $\begingroup$ Maybe $R\subseteq V_\kappa$? Also, do you mean why it is not satisfied under $\Pi^1_1$-describability? $\endgroup$ – Asaf Karagila May 12 '16 at 18:30
  • $\begingroup$ Have you tried to look at the proof of the equivalence to get some sense why these two things are equivalent? The intuition that we can give you is most likely "well, if this set is such and such ... " which is just an explanation of the proof which comes after the statement you cite. Do you understand the proof? $\endgroup$ – Asaf Karagila May 12 '16 at 18:34
  • $\begingroup$ Ok, I will try reformulate the question for a specific issue in the proof, thank you. $\endgroup$ – mikulas May 12 '16 at 19:04
  • $\begingroup$ Uhh, Where does the $++$ come from in the $V_{\alpha^{++}}$? Also, I'm not entirely clear. Which property seems "easy to satisfy", the extension or the indescribability? $\endgroup$ – Asaf Karagila May 12 '16 at 21:17
  • $\begingroup$ Sorry, another typo, that was meant to be "+1". I seems to me, that to construct an elementary extension for a given structure is easy. To do it with respect to any arbitrary predicate still seems easy, but obviously it's not since it's in fact a large cardinal property. Sorry I'm asking too vague questions, I'm trying to make sense of Kanamori's proof. Do you think it's what Erdős would call a "book proof"? To me, it doesn't make it clear even though I agree on the steps he takes, I believe it's a proof, but I still find it confusing that extension property is such a strong property. $\endgroup$ – mikulas May 13 '16 at 1:22

End-extensions are not at all trivial to obtain.

First of all, note that if $X$ is an end-extension of $V_\kappa$, then $X$ is a transitive model of $\sf ZFC$ of height $>\kappa$. This is not to be taken lightly.

Suppose $V=L$ and there exists a single inaccessible cardinal $\kappa$. Moreover assume that $\sf\operatorname{Con}(ZFC+I)$ is false in our universe. So there is no model of $\sf ZFC+I$ (where $\sf I$ is the existence of a strongly inaccessible cardinals). Now I claim that there is no end-extension of $V_\kappa$, since any end-extension of $V_\kappa$ will have to satisfy that $\kappa$ is a strongly inaccessible cardinal. And then you will have a model of $\sf ZFC+I$ (in fact a well-founded model!) which is impossible in our universe.

You might want to argue that the same would work for weak-compactness, but this is not the case. Weak-compactness is a second-order property over $V_\kappa$, and the end-extension only guarantees that there is a first-order end-extension.

So what does this mean? It means that if you have a weakly compact $\kappa$, then there are transitive models which have height greater than $\kappa$. This is not a trivial fact anymore.

Now you can easily prove that the end-extension property implies that $\kappa$ is at least strongly inaccessible, this is the first thing Kanamori proves (in the relevant direction). The idea is that a function $F\colon X\to\kappa$ which is cofinal satisfies that $\forall y\exists z(z\in X\land F(z)\notin y)$, as a first-order sentence over $V_\kappa$ (with $F$). But if you take an end-extension, this is not true anymore since now $\kappa$ is an element of the end-extension. This implies both regularity and being a strong limit.

So we see that not only the end-extension property implies strong inaccessibility. It is in fact much stronger than just inaccessibility.

  • $\begingroup$ A small remark about the strength of the extension - from the proof (of both directions) it seems that what you can get is not just a first-order end-extension, but an $L_{\kappa,\kappa}$ extension, which is much stronger (although not as strong as second order and still not strong enough to say that $\kappa$ itself is weakly compact). $\endgroup$ – Ur Ya'ar Feb 27 '18 at 11:07

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