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Again .

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.

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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 $i<\kappa$.

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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

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.

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Thank you very much – Marc Moretti May 3 '12 at 17:33

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