# proof of diamond

I have many questions about page 2 of this paper http://www.cs.elte.hu/~kope/ss3.pdf.

First, on the top, I try to prove that, if $cf(\delta)\neq\kappa$, then we can choose the $\alpha, \beta$ in an $A_i^\delta$.

1. We can prove that there exists an $i<\kappa$ so that the set $\{\delta<\lambda^+ : A_i^\delta \text{ is cofinal in }\delta\}$ is stationary in $\lambda^+$.
2. I know that, for all $i<\kappa$, the set $B_i^\delta=\{\alpha<\delta : \exists\beta<\delta, Z_i\cap\alpha=(X_\beta)_i\}$ is cofinal in $\delta$.
3. Intuitively, $cf(\delta)<\kappa$.
4. If $\delta$ is an ordinal then the cardinal of $\{X\subseteq\delta : X \text{ cofinal in } \delta\}$ is $2^\delta$ ?

but I can't "see" why we can take $\alpha,\beta$ in $A_i^\delta$.

Second, in Lemma 3, do we take a particular bijection $\pi$ ? I want to take the sequence $<\pi(X_\beta) : \beta<\lambda^+>$ but I have some difficulty to prove that $Z\cap(\lambda\times\alpha)=X_\beta$. I need $\pi(\alpha)=\lambda\times\alpha$...

Third, in the proof of the claim, I want to read $Y_\eta\cap\alpha=(X_\beta)_\eta$ instead of $Y_\eta=(X_\beta)_\eta$. Is it correct ?

Thanks.

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In the proof of Lemma 1 we have a $\delta<\lambda^+$ and a $\beta<\delta$ such that the set $$D=\left\{\alpha<\delta:\forall i<\kappa\Big(Z_i\cap\alpha=(X_\beta)_i\Big)\right\}$$ is cofinal in $\delta$. Moreover, $\delta\in S$, and by hypothesis $S\subseteq S_{\ne\kappa}^{\lambda^+}=\{\eta<\lambda:\operatorname{cf}\eta\ne\kappa\}$, so $\operatorname{cf}\delta\ne\kappa$. (This hypothesis is found in the statement of Theorem 1.) Fix a set $\Delta\subseteq\delta$ such that $\Delta$ is cofinal in $\delta$, and $|\Delta|=\operatorname{cf}\delta$.
Now recall that $\{A_i^\delta:i<\kappa\}$ is an increasing decomposition of $\delta$. This means that for each $\xi<\delta$ there is a smallest $i(\xi)<\kappa$ such that $\xi\in A_{i(\xi)}^\beta$. If $|\Delta|>\kappa$, there is an $i_0<\kappa$ such that $|\{\xi\in\Delta:i(\xi)=i_0\}|=|\Delta|$, and therefore $\{\xi\in\Delta:i(\xi)=i_0\}$ is cofinal in $\delta$. But then $A_i^\delta\cap\Delta$ is cofinal in $\delta$ for all $i$ such that $i_0\le i<\kappa$. If $|\Delta|<\kappa$, there is an $i_0<\kappa$ such that $i(\xi)<i_0$ for all $\xi\in\Delta$, and therefore $\Delta\subseteq A_i^\delta$ for all $i$ such that $i_0\le i<\kappa$. In either case choose any $i\ge i_0$ large enough so that $\beta\in A_i^\delta$, and you'll have arbitrarily large $\alpha<\delta$ with $\alpha,\beta\in A_i^\delta$.
In Lemma 3 it doesn't matter what bijection you use. That's the point of Remark 1: for any bijection $\pi$ from $\lambda^+$ onto $\lambda\times\lambda^+$, there is a club set $\Lambda$ in $\lambda^+$ such that for each $\delta\in\Lambda$, $\pi\upharpoonright\delta$ is a bijection of $\delta$ onto $\lambda\times\delta$. Just work with $\langle X_\beta:\beta\in\Lambda\rangle$; $\Lambda$ 'looks exactly like' $\lambda$ anyway.
Yes, I think that you're right that $Y_\eta=(X_\beta)_\eta$ should be $Y_\eta\cap\alpha=(X_\beta)_\eta$ in the proof of the Claim.
Thank you for your answer. At the begining, $\beta$ doesn't depend on $\alpha$ ? I thought that once you take arbitrary large an $\alpha<\delta$ then there exists a $\beta<\delta$ (in fact a $\beta_\alpha) so that ... – Marc Moretti May 8 '12 at 9:29 In your proof, you say :If$|\Delta|>\kappa$, there is an$i_0<\kappa$such that$|\{\xi\in\Delta:i(\xi)=i_0\}|=|\Delta|$, and therefore$\{\xi\in\Delta:i(\xi)=i_0\}$is cofinal in$\delta$. I don't see why : if$\delta$is a regular cardinal then it's ok but, here,$\delta$is only an ordinal less than$\lambda^+$. I will think about it ... – Marc Moretti May 8 '12 at 20:04 @Marc: As I read it,$\beta$does not depend on$\alpha$, but only on$\delta\$. – Brian M. Scott May 8 '12 at 20:06