# Are there uncountable cardinals $\kappa$ such that $|\kappa\cap\mathsf{Card}| = \kappa$?

All the cardinals $$\kappa\leq\aleph_0$$ have the property that there are precicely $$\kappa$$ cardinals less than $$\kappa$$. Of course, $$\aleph_1$$ lacks this property since there are only $$\aleph_0 +1= \aleph_0$$ cardinals less than it. And I suppose the same goes for all successor cardinals for the same reason.

But are there any limit cardinals for which the property returns? If so, are they in ZFC, or does one have to introduce large cardinals axioms to get them?

• Note that such cardinal must satisfy $\kappa=\aleph_\kappa$, and this property is called "an $\aleph$ fixed point". – Asaf Karagila May 6 at 10:47

## 1 Answer

Such cardinals do exist in ZF. Take your favourite ordinal $$\alpha$$ (the choice $$\alpha=0$$ is pedagogically popular) and construct the sequence $$\alpha_0:=\alpha,\,\alpha_{n+1}:=\omega_{\alpha_n}$$ for integers $$n\ge 0$$. We've specified $$\aleph_0$$ ordinals, and their union is an ordinal too, say $$\beta$$. It's a limit ordinal satisfying $$\beta=\omega_\beta$$, so there are $$|\beta|=\aleph_\beta$$ cardinals below $$\aleph_\beta$$.

• With little more work we can show that the class of these cardinals is closed and unbounded, which in some sense means that "the typical cardinal arbitrary has this property". – Asaf Karagila May 6 at 10:48
• @MichaelBevan As far as I know, only the one Asaf Karagila mentioned in a comment under your OP. – J.G. May 6 at 10:59
• Sorry, just deleted that comment by accident. Thanks! – Michael Bevan May 6 at 11:02
• Just a small typo in your notation: $\alpha_{n+1}$ does not depend on $\alpha_n$. – Mark Kamsma May 6 at 11:06
• @MarkKamsma Thanks; fixed. – J.G. May 6 at 11:09