# presaturated ideals

In this paper, Gitik and Shelah make the following claim (part of Proposition 1.5):

Claim (Gitik-Shelah): Suppose $\kappa < \lambda$ are regular, $2^\lambda = \lambda^+$, and $D$ is a normal ideal on $\lambda$. If forcing with $D^+$ makes $cf(\lambda^+) < \kappa$, then it collapses all cardinals in some interval $(\alpha,\lambda^+]$ where $\alpha < \kappa$.

I do not understand the argument for this claim. An interesting dichotomy follows:

Corollary: If GCH holds and $\eta$ is a singular cardinal, then for every normal ideal $D$ on $\eta^+$, either $D$ is presaturated or forcing with $D$ collapses $\eta$.

Sketch of proof: Assume $D$ is not presaturated but $D^+$ preserves that $\eta$ is a cardinal. Since $D$ is not presaturated, forcing with $D^{+}$ collapses $\eta^{++}$, and since $|D^+| = \eta^{++}$, it preserves $\eta^{+3}$. By a well-known theorem of Shelah, this implies that $cf(\eta^{++}) = cf(\eta) < \eta$ in the extension by $D^+$. But by the Claim, $\eta$ is collapsed, a contradiction.

However, it seems that we should be able to force a counterexample to this dichotomy from large cardinals, but I am probably missing some subtlety. So my question is, how do you prove the Claim?

• By Lemma 4.9 page 440 of Shelah's book proper forcing'', we must have in the generic extension, $cf(\lambda^+)=cf(|\lambda^+|)$. – Mohammad Golshani Feb 17 '15 at 4:28
• Yes, I am using this theorem in my corollary, but I don't see how it implies the claim. – Monroe Eskew Feb 17 '15 at 4:31
• Maybe a simple question in your argument. In your argument for applying Shelah's result, you seem use $\eta^{++}$ is collapsed into $\eta.$ Why this is true? in other words why $\eta^+$ is also collapsed? – Mohammad Golshani Feb 17 '15 at 5:16
• Simply because $\eta^+$ is the critical point of the generic embedding, so the ultrapower thinks that $\eta^+$ is an ordinal of cardinality $\eta$. In the non-precipitous case we use the fact that the generic ultrapower is well-founded at least up to $\eta^{++}$ (using canonical functions). – Monroe Eskew Feb 17 '15 at 5:18
• @Mohammad: Does this theorem appear in Proper and Improper Forcing somewhere as well? – Asaf Karagila Feb 17 '15 at 5:20

Theorem. Suppose $\kappa$ is a regular uncountable cardinal and $|P|\leq \kappa.$ Then $\Vdash_P cf(\kappa)=|\kappa|.$
In your case $D^+$ has size $\lambda^+$ and it forces $cf(\lambda^+)<\kappa.$ So by above theorem it also forces $|\lambda^+|<\kappa.$