Is the set of all cardinals smaller then a strongly inaccessible cardinal closed?

Given a strongly inaccessible cardinal $k$ (i.e. $k$ is regular uncountable and for each $\lambda < k$, $2^\lambda < k$), is the set of all cardinals smaller then $k$ closed or open?

Mahlo Cardinals

Let $\kappa$ be an inaccessible cardinal. The set of all cardinals below $\kappa$ is a closed unbounded subset of $\kappa$, and so is the set of its limit points, the set of all limit cardinals. In fact, the set of all strong limit cardinals below $\kappa$ is closed unbounded.
If $\kappa$ is the least inaccessible cardinal, then all strong limit cardinals below $\kappa$ are singular, and so the set of all singular strong limit cardinals below $\kappa$ is closed unbounded. If $\kappa$ is the $\alpha$th inaccessible, where $\alpha\lt\kappa$, then still the set of all regular cardinals below $\kappa$ is nonstationary.
An inaccessible cardinal $\kappa$ is called a Mahlo cardinal if the set of all regular cardinals below $\kappa$ is stationary.

-- Jech, Set Theory (3rd Millennium edition).

• I'm not quite sure from the context of the passage you've added, whether your question is what you were meaning to ask or not. – Asaf Karagila Nov 22 '14 at 14:04

The set of cardinals below $\kappa$ is always closed in $\kappa$ (it might not be closed if $\kappa$ is a limit cardinal).
The reason is that if $A$ is a set of cardinals, then $\sup A$ is a cardinal as well. So if $\sup A<\kappa$, it is a cardinal smaller than $\kappa$. Therefore $\{\lambda<\kappa\mid\lambda\text{ is a cardinal}\}$ is a closed set.
While we are at it, since you brought up limit cardinals, we can observe that if $\kappa$ is a limit cardinal, then whenever $\lambda<\kappa$, we have that $\lambda^+<\kappa$. Therefore there is no largest cardinal below $\kappa$. So in fact if $\kappa$ is a limit cardinal, the set of cardinals below it is a club and not just a closed set.
• I think that $k$ is a limit cardinal. Otherwise we would have $k=\alpha+1$ for some $\alpha < k$ where $|\alpha|=|k|$. I also think that the limit of a set of cardinals is a cardinal. This is why it seemed to me that the set describle cannot be closed because $k$ is not in it. – topsi Nov 22 '14 at 14:07