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I proved that a club $C$ in $\kappa$ has the same cardinality as $\kappa$. Is it really true ? Thanks.

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1 Answer 1

up vote 4 down vote accepted

Every club is an unbounded set. If $\kappa$ is regular this means that every unbounded set has order type $\kappa$ and therefore of size $\kappa$.

If $\kappa$ is singular then it is not true, take $\{\aleph_\alpha\mid\alpha<\omega_1\}$ as a club of $\aleph_{\omega_1}$.

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I supposed of course that $\kappa$ is regular .... thanks. –  Marc Moretti Apr 16 '12 at 15:01
    
@MarcMoretti: No problem. The proof of the claim "unbounded $\Rightarrow$ of order type $\kappa$" is simple. Suppose $A$ is unbounded in $\kappa$, write $A=\{a_i\mid i<\lambda\}$ and write $A_i = [a_i,a_{i+1})$ (the half-open interval) then $\bigcup A_i = \kappa$ and for all $i$ we have $|A_i|<\kappa$. Therefore $\lambda=\kappa$, as wanted. –  Asaf Karagila Apr 16 '12 at 19:15
    
ok. An other proof is this one : let $A\subset\kappa$ club. It is unbounded so its cardinality is greatest than the minimal cardinality of a cofinal subset of $\kappa$, say $cf(\kappa)$. So $cf(\kappa)\leq|A|\leq\kappa$. But $\kappa$ regular so $|A|=\kappa$ as wanted. In fact, this proof is ok for just unbounded subset of $\kappa$. Regards. –  Marc Moretti Apr 18 '12 at 8:55
    
@Marc: Your proof is good, better than mine. As you noticed, indeed the proof only required unboundedness and not closure of any kind. –  Asaf Karagila Apr 18 '12 at 8:59

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