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In a paper, I want to prove a result that seems to me general.

Let $g:\delta\longrightarrow cf(\lambda)$ where $\delta$ is an ordinal less than $\lambda^+$ and $\lambda$ a cardinal. Suppose that $\forall i<cf(\lambda)$, $g^{-1}[i]$ is not cofinal in $\delta$. Does one have $cf(\lambda)=cf(\delta)$ ?

Remark : I think the notation $g^{-1}[i]$ means the inverse image that is $\{\alpha<\delta : g(\alpha)<i\}$.

We have $\delta=g^{-1}[cf(\lambda)]=\bigcup_{i<cf(\lambda)}g^{-1}[i]=\sup_{i<cf(\lambda)} \delta_i$ where $\delta_i$ is the least upper bound of $g^{-1}[i]$ in $\delta$ so $cf(\delta)\leq cf(\lambda)$. I have some difficulty for the other direction. Could somebody help me ?


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In your first line, do you intend to prove this in a paper, or do you read a paper and this is the stated result? – Asaf Karagila Jun 9 '12 at 21:58
@Asaf : I 've read a result in a paper and to prove it I need to prove the one I wrote. – Marc Moretti Jun 10 '12 at 7:07
up vote 2 down vote accepted

Let $\kappa=\operatorname{cf}\delta$, and let $\langle\alpha_\xi:\xi<\kappa\rangle$ be a cofinal sequence in $\delta$. For each $\beta<\operatorname{cf}\lambda$ there is a $\xi(\beta)<\kappa$ such that $g(\alpha_{\xi(\beta)})\ge\beta$, since $g^{-1}[\beta]$ is not cofinal in $\delta$. If $\kappa<\operatorname{cf}\lambda$ there is a $\xi<\kappa$ such that $B=\{\beta<\operatorname{cf}\lambda:\xi(\beta)=\xi\}$ is cofinal in $\operatorname{cf}\lambda$. But then $g(\alpha_\xi)\ge\sup B=\operatorname{cf}\lambda$, which is impossible. Thus $\kappa\ge\operatorname{cf}\lambda$, and since you already have $\kappa\le\operatorname{cf}\lambda$, it follows that $\operatorname{cf}\delta=\kappa=\operatorname{cf}\lambda$.

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Thanks a lot. It is a classical proof that I need to work. – Marc Moretti Jun 10 '12 at 7:11

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