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I wanted to prove that if $C$ is a club (in a $\kappa$ cardinal) and $C\subseteq C'$ then $C'$ is also a club (that is if $C$ is of "mesure" $1$ then $C'$ is too). It is easy to see that $C'$ is unbounded. But I have some problems proving the closeness of $C'$. So, I try to find a $C'$ that is not a club!

Could somebody help me? thanks.

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2 Answers 2

up vote 4 down vote accepted

First observe that it is true for any club $C$ such that $\mathrm{Lim}\cap\kappa\subseteq C$, since $C$ already contains all the limit points possible.

However if $\delta<\kappa$ is a limit ordinal and $\delta\notin C$ we can take $C'=C\cup\delta$, then in $\delta$ we have that $\sup(C'\cap\delta)=\delta\notin C'$, so it is not a club.

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$C'$ need not to be closed. Consider $C = \{\omega \alpha\mid 2\le \alpha < \omega_1\} \subseteq \omega_1$ which is club in $\omega_1$. But $C' := \omega \cup C$ isn't closed.

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what do you mean by $\omega\alpha$ ? –  Marc Moretti Jul 4 '12 at 7:37
    
@Marc: Ordinal multiplication, this is the $\alpha$-th limit ordinal. –  Asaf Karagila Jul 4 '12 at 7:40
    
@Asaf : thanks for your answer. –  Marc Moretti Jul 4 '12 at 10:48

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