Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Again http://www.cs.elte.hu/~kope/ss3.pdf .

After the Remark 2, I have some problem to prove that there exists a increasing decomposition of $\delta<\lambda^+$ ($\delta$ ordinal ? or cardinal ?) as $$\delta=\bigcup_{i<\kappa}A_i^\delta$$ with $|A_i^\delta|<\lambda$. I've tried to distinguish according to the position of $\delta$ regard to $cf(\lambda)$ and $\lambda$ etc... Another question : do we implicitly suppose that $\lambda$ is a cardinal ? because, $\lambda^+$ is defined to be the first cardinal greater than the ordinal $\lambda$ ... ?

Could somebody give me an indication ? Thanks.

share|improve this question
add comment

2 Answers 2

up vote 1 down vote accepted

The assumptions here are those of Theorem 1: $\lambda$ is a cardinal, $2^\lambda=\lambda^+$, and $\kappa=\operatorname{cf}(\lambda)$. Since $\operatorname{cf}(\lambda)=\kappa$, there is an increasing $\kappa$-sequence $\langle \eta_\xi:\xi<\kappa\rangle$ of ordinals less than $\lambda$ such that $\sup\{\eta_\xi:\xi<\kappa\}=\lambda$. As a result, $\lambda$ can be written as the union of the $\kappa$ sets $[0,\eta_\xi)$, each of which has cardinality less than $\lambda$.

In the comment after Remark 2, $\delta$ ranges over all ordinals less than $\lambda^+$. In particular, $|\delta|\le\lambda$. If $|\delta|=\lambda$ (i.e., if $\lambda\le\delta<\lambda^+$), use a bijection between $\delta$ and $\lambda$ to write $\delta$ as an union of $\kappa$ sets, each of cardinality less than $\lambda$. If $\delta<\lambda$, just let $A_i^\delta=\delta$ for all $i<\kappa$.

share|improve this answer
    
Thank you. I thought that it was a general result and not with the assumptions of the theorem. –  Marc Moretti May 3 '12 at 17:32
    
@Marc: I've not yet had time to read the paper thoroughly, but at a quick glance it appears to me that that the hypotheses of Theorem 1 are maintained throughout. –  Brian M. Scott May 3 '12 at 17:36
add comment

At least in Shelah's work $\lambda$ is almost always used to denote a cardinal.

We have, in Remark 2, that $\lambda$ is a cardinal of cofinality $\kappa$ and $\delta$ is an ordinal between $\lambda$ and $\lambda^+$. So $\delta$ has size $\lambda$ so it has a partition into $\kappa$ many increasing sets since $\lambda$ has such partition.

share|improve this answer
    
Thank you very much –  Marc Moretti May 3 '12 at 17:33
add comment

Your Answer

 
discard

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.