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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).

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  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Commented Nov 22, 2014 at 14:04

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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.

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  • $\begingroup$ The problem is, that answering these two questions leads me to the conclusion that this set is not closed. But in Jech's book it is written that it is closed... $\endgroup$
    – topsi
    Commented Nov 22, 2014 at 13:56
  • $\begingroup$ Well, how did you answer them? $\endgroup$
    – Asaf Karagila
    Commented Nov 22, 2014 at 14:01
  • $\begingroup$ "closed" could mean different things, closed set in a topological space, but also closed under a set of algebraic of other operations. E.g. the complex numbers are closed in the latter sense. The transitive closure (in set theory) is closed too (even if we do not talk about topology here). So try to figure in what sense the author meant that the set in questions is closed. I may be off, but this is something to consider. $\endgroup$
    – Mirko
    Commented Nov 22, 2014 at 14:06
  • $\begingroup$ 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. $\endgroup$
    – topsi
    Commented Nov 22, 2014 at 14:07
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    $\begingroup$ @user48481MirkoSwirko: In the order topology on ordinals, being closed topologically means being closed under taking supremums. $\endgroup$
    – Asaf Karagila
    Commented Nov 22, 2014 at 14:09

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