2
$\begingroup$

in definition of $\alpha$-inaccessible cardinals on Wikipedia, we can read :

For example, denote by $ψ_0(λ)$ the λth inaccessible cardinal, then the fixed points of $ψ_0$ are the 1-inaccessible cardinals.

  1. But, the function $\psi_0$ is not a normal function (as it would implies that $cf(\psi_0(\omega))=\omega$, and, not so inaccessible !). So what (smallest, if any) hypothesis do we need to prove that $\psi_0$ has a fixed point and how to ?

  2. same question for $\alpha$-hyper-inaccessible.

  3. Is it sufficient for a cardinal to be inaccessible, hyper-inaccessible, hyper-hyper-inaccessible,... and so on, to be Mahlo ?

$\endgroup$
1
$\begingroup$

You don't need the normality in order to prove that $1$-inaccessible are the fixed points. It's in fact trivial from the fact that inaccessible cardinals are regular.

  • $\kappa$ is $1$-inaccessible if and only if
  • $\kappa$ is a limit of inaccessible cardinals if and only if
  • $\kappa$ has $\kappa$ inaccessible below it if and only if
  • $\kappa$ is the $\kappa$-th inaccessible.

The same point goes for the second question, there you can replace $\psi_0$ by $\psi_\alpha$ to argue for $\alpha+1$. For limit ordinals it follows from the definition, too.

Finally, you can find the answer to the last question here: mahlo and hyper-inaccessible cardinals

$\endgroup$
  • $\begingroup$ Thank you ! So we only need to suppose there are as many inaccessible cardinals as we want ? Or can we suppose less ? $\endgroup$ – Xoff Aug 27 '13 at 15:28
  • $\begingroup$ If it's for fun, then sure. You can suppose there are many, and you can suppose that there are less. If it's for something else, then you might want to let me know before I can answer. :-) $\endgroup$ – Asaf Karagila Aug 27 '13 at 15:29
  • $\begingroup$ In fact, I would like to understand : if I suppose there is one inaccessible cardinal, does it imply there is one 1-inaccessible as well ? and one 1-hyper-inaccessible ? $\endgroup$ – Xoff Aug 27 '13 at 15:35
  • 1
    $\begingroup$ @Xoff: No. Of course not. The list is just a way to unfold the definitions. $\kappa$ is $2$-inaccessible if it is the $\kappa$ $1$-inaccessible; and so on. $\endgroup$ – Asaf Karagila Aug 27 '13 at 15:40

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.