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I try to prove the equivalence between "C is closed in an ordinal $\alpha$" and "each strictly increasing sequence of elements of C of length $< cf\alpha$ converge in C".

For $\Rightarrow$, it's ok.

For $\Leftarrow$ : let $\gamma<\alpha$ ordinal limit such that $Sup(C\cap\gamma)=\gamma$. I want to show that $\gamma\in C$. Let $\beta<\gamma$. There exists $c_0\in C$ such that $\beta<c_0<\gamma$ so $c_0\in C\cap\gamma$. Suppose $c_\xi\in C\cap\gamma$ constructed and define $c_{\xi+1}$ as we done with $c_0$. My questions are :

1) what is the length $\gamma'$ of this sequence ?

2)Do we have $\gamma'<cf\alpha$ ?

3) Does each strictly increasing sequence of ordinals $<\alpha$ is of length $< cf\alpha$ ?


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Are you sure you mean strictly increasing sequences of length $< \mathrm{cf} \alpha$? (Consider the family of all successor ordinals $< \omega_\omega$.) – arjafi Nov 17 '12 at 18:20
oups. I meant strictly increasing and bounded in $\alpha$. – Marc Moretti Nov 17 '12 at 20:11
up vote 1 down vote accepted
  1. The length of the sequence $c_\xi$ is at least as the cofinality of $\gamma$. Note that $\gamma$ is pretty much any limit ordinal, so it may be any cofinality. Even countable.

  2. Consider the ordinal $\alpha=\omega_2+\omega_1$, and a closed $C$ in $\alpha$ which reflects to $\omega_2$ as a club, namely $C\cap\omega_2$ is closed and unbounded in $\omega_2$, it follows that $\omega_2\in C$, but the cofinality of $\alpha$ is only $\omega_1$.

  3. Of course not, in the case above consider the increasing sequence $\gamma<\omega_2$ below $\alpha$, whose cofinality is $<\omega_2$.

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so, my result is only for $\alpha$ regular ? – Marc Moretti Nov 17 '12 at 20:43
@Marc: Yes. You should replace the "sequence of length $<\operatorname{cf}(\alpha)$" with "bounded sequence". – Asaf Karagila Nov 17 '12 at 20:45

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