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Good evening.

Let $\lambda>\omega$ a cardinal. We know that there is a bijection $\pi$ between $\lambda^+$ and $\lambda\times\lambda^+$. I don't understand in Remark 1 of the paper Shelah's proof of diamond of Komjath the words : "a club set of $\delta<\lambda^+$ (do we take a $C\subset\delta$ club in $\delta$ ? or ... ?)" $\pi$ is a bijection from $\delta$ onto $\lambda\times\delta$ " (why ?) Thanks a lot.

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Please add a citation of the paper you read, and link if possible! –  Asaf Karagila Apr 27 '12 at 12:44
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up vote 3 down vote accepted

The paper is here. The actual remark is:

If $\pi$ is a bijection from $\lambda^+$ onto $\lambda\times\lambda^+$ then for a club set of $\delta<\lambda^+$ (the restriction of) $\pi$ is a bijection from $\delta$ onto $\lambda\times\delta$.

It means that $D\triangleq\{\delta<\lambda^+:\pi\upharpoonright\delta\text{ is a bijection onto }\lambda\times\delta\}$ is a club set in $\lambda^+$. Verification that $D$ is closed is straightforward, and verification that $D$ is unbounded is almost as easy: just use the usual closing-off construction.

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Thank you for your answer ! Regards –  Marc Moretti Apr 27 '12 at 13:09
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