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Let $\kappa$ be an infinite cardinal.

$\kappa$ is stable under ordinal addition $+$, ordinal multiplication $.$ and ordinal exponentiation $e: (a,b) \mapsto a^b$, so $\mathcal{K} = (\kappa,+,.,e,\in)$ is a model theoritic structure.

a) Is the set of cardinals $< \kappa$ first order definable in $\mathcal{K}$?

b) Which cardinals $< \kappa$ are first order definable in $\mathcal{K}$?

Note that if $\{|\lambda| \ | \ \lambda \in \kappa\}$ is definable then so are all $\aleph_{\alpha}, \alpha < 2.\omega_0$ (provided they are in $\lambda$), and you can probably go much further.

Not too much however, for one can only define countably many cardinals, so there is a least undefinable cardinal $\leq \aleph_{\omega_1}$.

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a) Note that $\mathcal K =(\kappa, +, e, \in)$ is absolute between transitive models of ZFC. Thus any collapse below $\kappa$ demonstrates that the set of cardinals below $\kappa$ cannot be first order definable in $\mathcal K$ if $\kappa > \omega_1$.

b) Sorry, I'm not sure how to prove my initial claim that the answer is $\omega + 1$. I will add to this answer in a moment.


Actually, the situation is easier: If $\kappa$ is uncountable, collapse $\kappa$ to $\omega_1$. As $(\omega_1,+,e,\in)$ cannot define uncountable cardinals, absoluteness yields the result.


To clearify: What this answer actually says: Let $\phi(x)$ be a $\{+,e,\in\}$-formula and let $\kappa$ be an ordinal. We cannot prove (in ZFC) that $$\phi( (\kappa,+,e,\in) ) := \{ \alpha < \kappa \mid (\kappa,+,e,\in) \models \phi(\alpha) \} $$ contains a cardinal above $\omega$.

On the other hand, let $$ \phi(x) = x \in \omega \vee x = \omega \vee \forall y \colon y = x \vee y \in x $$

Then $\phi((\omega_1+1; \in)) = \omega+1 \cup \{\omega_1\}$ and thus $\phi(x)$ defines a set of cardinals, that contains an uncountable cardinal. This may seem very strange to you, but what my answer basically comes down to: Above $\omega$, ZFC tells us nothing about which ordinals are cardinals... (Even if we restrict our attention to transitive models with "the true" membership relation.)

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    $\begingroup$ Thanks for your post. I lack the knowledge necessary to understanding it but I will keep it in mind and think about it again when I figure I can. $\endgroup$ – nombre Jul 2 '15 at 16:05
  • $\begingroup$ Well, this approach uses blunt force to show that $(\kappa,+,e,\in)$ may now very little about its cardinals. It's definitely not elegant, but it came to mind and as this question hasn't had any responses yet, I figured that I should share my thoughts. $\endgroup$ – Stefan Mesken Jul 2 '15 at 16:25
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    $\begingroup$ Ok I think I get it. ($\phi(\omega_1 + 1,\in)$ should rather be $\omega \cup \{\omega_1\}$ though) $\endgroup$ – nombre Jul 2 '15 at 18:26
  • $\begingroup$ Actually no, but I probably should've written it down in a different fashion: $\omega +1 = \omega \cup \{\omega \}$, so $\omega+1 \cup _\{omega_1\}$ consists of all finite ordinals, $\omega$ and $\omega_1$ (nothing more). Sorry if I unintentionally confused you by using ordinal addition instead of unions. $\endgroup$ – Stefan Mesken Jul 2 '15 at 19:21
  • $\begingroup$ Oh, there was a typo in my formula. Sorry. $\endgroup$ – Stefan Mesken Jul 2 '15 at 19:23

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