Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Please add a citation of the paper you read, and link if possible! – Asaf Karagila Apr 27 '12 at 12:44
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.

share|cite|improve this answer
Thank you for your answer ! Regards – Marc Moretti Apr 27 '12 at 13:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.