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Let $\alpha$ an ordinal and $\langle\alpha_\xi\rangle$ a cofinal sequence of elements of $\alpha$. The length, $\gamma$, of this sequence is at least $\operatorname{cf}\alpha$ but can be equal to any ordinal $\geq \operatorname{cf}\alpha$. If this sequence is a strictly increasing one then $\gamma$ is $\leq$ to the order type of $\alpha$ that is $\alpha$. Is it exact ?

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up vote 3 down vote accepted

If $\alpha$ is a limit ordinal and $\langle\alpha_\xi:\xi<\gamma\rangle$ is a non-decreasing sequence cofinal in $\alpha$, then $\operatorname{cf}\gamma=\operatorname{cf}\alpha$, even if the sequence is not strictly increasing. As you say, if the sequence is strictly increasing, then $\gamma\le\alpha$.

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Thanks. I forgot Jech, set theory p.32 ... –  Marc Moretti Nov 18 '12 at 21:49
    
@Marc: You’re welcome. –  Brian M. Scott Nov 18 '12 at 21:50

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