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Definition 1. An uncountable cardinal $\kappa$ is Shelah if for every function $f:\kappa\rightarrow \kappa$ there exists a transitive class $M$ and a non-trivial elementary embedding $j:V\rightarrow M$ such that $^\kappa M\subseteq M$, $crit(j)=\kappa$ and $V_{j(f)(\kappa)}\subseteq M$.

Definition 2. Woodin's fast function forcing on $\kappa$ consists of partial functions $p$ from $\kappa$ to $\kappa$ ordered by inclusion such that:

  • The domain of $p$ consists of inaccessible cardinals $\lambda <\kappa$ which are closed under $p$, namely for every $\lambda, \theta\in dom(p)$ if $\theta<\lambda$ then $p(\theta)<\lambda$.

  • For every $\lambda\in dom(p)$ we have $|dom(p)\cap \lambda|<\lambda$

Remark. Following Joel's comment below, it is worth mentioning that the above definition is NOT the only variant of fast function forcing. There are other versions with slightly different properties. For a more complete argument along these lines see Joel's answer in this MO post.

On one hand, preserving Shelah cardinals through lifting arguments often needs dominating the corresponding functions $f:\kappa\rightarrow \kappa$ by functions in the ground model.

On the other hand, we know that the fast function forcing $\mathbb{P}_{\kappa}$ adds a very fast (and so non-dominatable) function of this type into the universe (i.e. the fast function) and simultaneously fails to satisfy $\kappa$-cc property which is a usual condition for providing dominating functions in the ground model.

Thus it is somehow natural to expect that (all variants of) Woodin's fast function forcing can kill Shelah cardinals. Is it true? Any concrete example?

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    $\begingroup$ The fast function forcing works better, with nicer factor properties, if you insist that $|p\upharpoonright\lambda|<\lambda$ for every inaccessible cardinal, rather than just $\lambda\in\text{dom}(p)$. $\endgroup$ – Joel David Hamkins Dec 7 '17 at 1:42
  • $\begingroup$ @JoelDavidHamkins Thanks for reminding this fact, Joel! I edited the original post accordingly. $\endgroup$ – Morteza Azad Dec 7 '17 at 6:34
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    $\begingroup$ What about the following idea: Assume it preserve the Shelahness of $\kappa$, where $\kappa$ is the least Shelah cardinal. Let $f: \kappa \to \kappa$ be the fast function added by the forcing $\mathbb{P}$. Let $G$ be the generic filter. By our assumption, there exists $j: V[G] \to M$ such that $M$ is closed under $\kappa$-sequences (in $V[G]$), $crit(j)=\kappa$ and $V[G]_{j(f)(\kappa)} \subset M.$ Let us force below $(\kappa, wt^V(\kappa)+\omega) \in j(\mathbb{P}),$ where $wt^V(\kappa)$ is the witnessing number of $\kappa$ in $V$. $\endgroup$ – Mohammad Golshani Dec 7 '17 at 8:18
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    $\begingroup$ This implies $V[G]_{wt^V(\kappa)+\omega} \subset M,$ in particular, the results of An Easton like theorem in the presence of Shelah cardinals show that $\kappa$ is not the least Shelah cardinal in $V[G]$. But the forcing creates no new Shelah cardinals, so $\kappa$ is not the least Shelah cardinal in $V$ as well, a contradiction. $\endgroup$ – Mohammad Golshani Dec 7 '17 at 8:19
  • $\begingroup$ @MohammadGolshani Thanks! Your approach sounds promising! Could you please add it as an answer? $\endgroup$ – Morteza Azad Dec 7 '17 at 9:58
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Assume it preserve the Shelahness of $\kappa$, where $\kappa$ is th eleast Shelah cardinal. Let $f: \kappa \to \kappa$ be the fast function added by the forcing $\mathbb{P}$. Let $G$ be the generic filter.

By our assumption, there exists $j: V[G] \to M$ such that $M$ is closed under $\kappa$-sequences (in $V[G]$), $crit(j)=\kappa$ and $V[G]_{j(f)(\kappa)} \subset M.$

Let us force below $(\kappa, wt^V(\kappa)+\omega) \in j(\mathbb{P}),$ where $wt^V(\kappa)$ is the witnessing number of $\kappa$ in $V$. This implies $V[G]_{wt^V(\kappa)+\omega} \subset M,$ in particular, the results of An Easton like theorem in the presence of Shelah cardinals show that $\kappa$ is not the least Shelah cardinal in $V[G]$. But the forcing creates no new Shelah cardinals, so $\kappa$ is not the least Shelah cardinal in $V$ as well, a contradiction.

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  • $\begingroup$ (+1) Sounds correct! Thanks for posting, Mohammad! $\endgroup$ – Morteza Azad Dec 8 '17 at 6:12
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    $\begingroup$ I don't understand the argument: do you start by assuming that $j$ is some embedding in $V[G]$ or do you assume that this embedding extends an embedding from $V$? In the first case, you might not be able to pick $j(f)(\kappa)$ to be any value that you want. It shouldn't be an issue that the second case always occurs. The forcing admits a gap below $\kappa$ so any elementary embedding in $V[G]$ extends one from the ground model. $\endgroup$ – Yair Hayut Dec 10 '17 at 6:27
  • $\begingroup$ I think Hamkins work shows that any embedding in the extension is extension of some embedding from the ground model $\endgroup$ – Mohammad Golshani Dec 10 '17 at 13:18

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