# 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.

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

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

3. What upper and lower bounds can we obtain for $\Omega(\kappa)$ using the ZFC axioms?

• You may want to specify that $\kappa'$ is a cardinal itself. Apr 30, 2013 at 16:02
• Let $\kappa_0=\omega$. Given $\kappa_n$ for $n\in\omega$, let $\kappa_{n+1}=\omega_{\kappa_n}$. Let $\lambda=\sup\{\kappa_n:n\in\omega\}$; then $\Omega(\lambda)=\lambda$. Apr 30, 2013 at 16:03

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 $$\Omega(\kappa)=\kappa$$.

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