Order type of the set of all cardinals strictly below any given cardinal For any cardinal number $\kappa$, write $\Omega(\kappa)$ for the unique ordinal that is order isomorphic to $\{\kappa'\mid\kappa' < \kappa, \kappa'\text{ is a cardinal}\}.$ So for finite $\kappa$, we have that $\Omega(\kappa)=\kappa.$ This also holds when $\kappa = \omega$. Furthermore, we have that $\Omega(\aleph_\alpha)=\omega+\alpha,$ and that $\Omega(\beth_\alpha)\geq\omega+\alpha.$ Also, GCH is equivalent to the statement that $\Omega(\beth_\alpha)=\omega+\alpha$ for all $\alpha$.
A few questions come to mind.


*

*Does ZFC prove that there exist cardinal numbers $\kappa$ such that $\Omega(\kappa) \neq \kappa$?

*Does ZFC prove that there exist cardinal numbers $\kappa > \omega$ such that $\Omega(\kappa) = \kappa$?

*What upper and lower bounds can we obtain for $\Omega(\kappa)$ using the ZFC axioms?
 A: Before discussing the answers let us talk about something which is called an $\aleph$ fixed point. An infinite cardinal $\kappa$ is an $\aleph$ fixed point if $\kappa=\aleph_\kappa$. That is to say that the infinite cardinals below $\kappa$ have exactly order type $\kappa$.
To prove such cardinal exists, let $\kappa_0$ be any infinite cardinal and let $\kappa_{n+1}=\aleph_{\kappa_n}$. Consider $\kappa=\sup\{\kappa_n\mid n\in\omega\}$. We want to show that $\kappa=\aleph_\kappa$. Note that $\aleph_\kappa\geq\kappa$ is trivial. In the other direction note that if $\alpha<\kappa$ then $\alpha<\kappa_n$ for some $n$ and therefore $\aleph_\alpha<\kappa_{n+1}<\kappa$. Therefore for all $\alpha<\kappa$ we have $\aleph_\alpha<\kappa$ and the equality ensues.
The result is a limit cardinal which is singular and have countable cofinality. We may proceed further and produce fixed points of every possible cofinality as well.
One last remark is that if $\kappa$ is such fixed point then $\kappa$ must be an uncountable ordinal, and therefore when considering ordinal addition, $\omega+\kappa=\kappa$. This shows that for a fixed point $\Omega(\kappa)=\kappa$.
Now to answer your questions three.

*

*Of course. $\kappa=\aleph_1$ has the property that $\Omega(\omega_1)=\omega+1<\omega_1$.


*Again, the answer is positive, as seen above.


*Since for every $\lambda$ there is a fixed point $\kappa>\lambda$, we have that we can prove $\Omega(\kappa)$ can be as large as we want it. As for lower bound, obviously $0$ for the empty set. But for infinite cardinals I suppose that if $\Omega(\kappa)=\lambda$ then $\Omega(\kappa^+)=\lambda+1$ (ordinal addition).
So if $\kappa$ is any infinite cardinal $\Omega(\kappa)$ is at least $\omega$, and if it is uncountable then $\omega+1$.
