A cardinal $\kappa$ is weakly Mahlo if it's weakly inaccessible and the set $\{\lambda\in\kappa:\,\lambda\,\text{weakly inaccessible}\}$ is stationary in $\kappa$.

Let's define $E_0=\{\lambda:\,\lambda\,\text{weakly inaccessible}\}$, $E_{\alpha+1}=\{\lambda\in E_\alpha:\,|\lambda\cap E_\alpha|=\lambda\}$ and $E_\lambda=\bigcap_{\alpha<\lambda}E_\alpha$ if $\lambda$ is a limit ordinal. Now, I'm interested in see that $\kappa\in E_\kappa$ and, in fact, that $\kappa$ is the $\kappa^{th}$-cardinal of $\{\alpha:\alpha\in E_\alpha\}$. Does anybody have any idea how to see this?

Thanks in advance.

  • $\begingroup$ I've already reduced the problem to prove the following fact. Given $\alpha\in \kappa$, does it true that $\{\mu\in\kappa:(\mu\cap E_\alpha)\,\text{unbounded in}\,\mu\}$ is unbounded in $\kappa$? $\endgroup$
    – Cesare
    Jun 2, 2016 at 20:23

1 Answer 1


It is enough to show that actually $E_\alpha\cap\kappa$ is stationary by induction on $\alpha<\kappa$.

The case $\alpha=0$ follows from the fact that $\kappa$ is weakly inaccesible, and thus the set of limit cardinals less than $\kappa$ is closed unbounded.

There are two cases:

  1. Suppose $E_\alpha\cap\kappa$ is stationary. Then $(E_\alpha\cap\kappa)'$ is club and thus $(E_\alpha\cap\kappa)'\cap E_\alpha$ is stationary, and as this is a subset of $E_{\alpha+1}$, we get that $E_{\alpha+1}$ is stationary.

  2. Now suppose $\alpha<\kappa$ is limit, and all $E_\beta$ are stationary for $\beta<\alpha$. Suppose $E_\alpha$ is not stationary. Then there is a club $C$ such that for each $\beta\in C$ there is some $\xi<\alpha$ with $\beta\notin E_\xi$. Let, for each $\beta\in C$, $f(\beta)$ be the least $\xi$ such that $\beta\notin E_\xi$; notice that each $f(\beta)$ is a successor ordinal and $f(\beta)<\alpha.$ There exists a stationary set $S\subseteq C\cap E_0$ such that $f$ is constant on $S$, with value $\xi+1$. Pick some $\alpha\in (E_\xi\cap\kappa)'\cap S$; $(E_\xi\cap\kappa)'$ is club since $E_\xi\cap\kappa$ is stationary , then $\alpha\in E_\xi$ as $\alpha\in S$ because $f(\alpha)=\xi+1$. Then as $\alpha$ is also a regular cardinal; $S\subseteq E_0$, we get that since $\alpha\in (E_\xi\cap\kappa)',$ $|\alpha\cap E_\xi|=\alpha$, therefore $\alpha\in E_{\xi+1}$. Contradiction.

Hence as $E_\alpha\cap\kappa$ is stationary for each $\alpha<\kappa$, we get $\kappa\in E_\kappa$.

Now let us see that in fact $\{\alpha<\kappa:\alpha\in E_\alpha\}$ is stationary. Suppose not, then there is a club $C$ such that for each $\alpha\in C$ there is some $\xi<\alpha$ with $\alpha\notin E_\xi$. Define $f:C\rightarrow \kappa$ as in (2), then arguing similarly there is a stationary set $S\subseteq C\cap E_0$ such that $f$ has constant value $\xi+1$, and we get a similar contradiction.

The argument I had initially given, that is, to use that there is a stationary set $S\subseteq\kappa$ of weakly inaccesibles such that $V_\alpha\preceq V_\kappa$ for all $\alpha\in S$, is shorter but only works if $\kappa$ is also inaccesible though.

  • $\begingroup$ Thank you @CamiloArosemena by your answer. I'm willing to prove that $A=\{\alpha\in\kappa:\;V_\alpha\prec V_\kappa\}$ is a club and thus, by weakly Mahloness of $\kappa$, that the set of weakly inaccessibles in $A$ form a stationary set. But, could you be more explicit regarding how to use it to see $\kappa\in E_\alpha$? $\endgroup$
    – Cesare
    Jun 3, 2016 at 9:14
  • $\begingroup$ @Cesare, the answer I gave before was wrong, I just corrected my answer. $\endgroup$ Jun 3, 2016 at 17:11
  • $\begingroup$ Thank you @CamiloArosemena. But let me ask you only two more things: 1- Does $(E_\xi\cap \kappa)'$ means the set of cardinals below $\kappa$ which are limit of cardinals in $E_\xi\cap \kappa$? Or, in other words, does that comma refers to the Cantor-Bendisxon derivative? 2-How could work the argument you posted first for inaccessible ones? It's clear to me that this argument can be applied to show the same in the inaccessible setting. Anyway, could you show me why when $\{\alpha\in\kappa:\;V_\alpha\prec V_\kappa\}$ is stationary then $\kappa\in E_\kappa$? $\endgroup$
    – Cesare
    Jun 3, 2016 at 17:54
  • $\begingroup$ Regarding to the fact of being the $\kappa^{th}$-element of $A=\{\alpha\in\kappa:\alpha\in E_\alpha\}$, does it mean that $\kappa$ is the least cardinal for which $|A\cap\kappa|=\kappa$? $\endgroup$
    – Cesare
    Jun 3, 2016 at 17:57
  • $\begingroup$ @Cesare, $(E_\xi\cap\kappa)'$ is the set of limit points of $E_\xi\cap\kappa$ in $\kappa$. $\endgroup$ Jun 3, 2016 at 22:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .