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Suppose that $\kappa$ is an appropriate large cardinal (preferably a Woodin cardinal, but possibly something stronger) and let $G$ be a $\operatorname{Col}(\omega_1,<\kappa)$-generic filter over $V$. Then does $V[G]$ satisfy the following?

For a generic ultrafilter $H\subseteq \mathrm{NS}^+_{\omega_1}$ over $V[G]$ and a function $f\colon [\omega_1]^{<\omega}\to 2$ in $V[G]$, there is a set $S\in H$ such that $f$ is homogeneous over $S$.

Or can we force the above statement from $\mathsf{ZFC}$ with some large cardinal axiom?


The motivation for this question is whether the following variant of Prikry forcing can have the Prikry property:

$\mathbb{P}$ is the set of all $(a,S)$ where $a$ is an finite sequence over $\omega_1$ and $S$ is a stationary subset of $\omega_1$. $(b,T)\le (a,S)$ if $b$ is an end extension of $a$, $T\subseteq S$, and $b\setminus a\subseteq S$.

This forcing may not preserve cardinals larger than $\omega_2$ (the usual argument for the Prikry forcing over a measurable $\kappa$ is $\kappa$-c.c. does not work due to the extra condition $T\subseteq S$ in the definition of $\le$), but it seems that the above Ramsey-like property for $\mathrm{NS}^+_{\omega_1}$ guarantees $\mathbb{P}$ has a version of Prikry property, so $\mathbb{P}$ does not add bounded countable subsets of $\omega_1$.

I am more interested in the case when $\omega_1$ is replaced by another regular cardinal, and the most interesting case for me is $\omega_{\omega+1}$, which might give another way to singularize $\omega_{\omega+1}$ with adding no bounded subsets of $\omega_{\omega+1}$.

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  • $\begingroup$ Do you mean $f:[\omega_1]^{<\omega} \to 2$? $\endgroup$ Commented Feb 20 at 23:25
  • $\begingroup$ @MonroeEskew Yes, fixed. $\endgroup$
    – Hanul Jeon
    Commented Feb 20 at 23:26
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    $\begingroup$ This seems impossible, since the forcing adds a cofinal $\omega$ sequence in $\omega_1$, so it adds reals. But also why would the Prikry property imply no new reals? The intersection of countably many stationary sets can be empty. $\endgroup$ Commented Feb 20 at 23:28
  • $\begingroup$ @MonroeEskew My argument for the 'Prikry property' is that from a given condition $(a,S)$ and a forcing statement $\phi$, define a function $f\colon [\omega_1]^{<\omega}\to 2$ as in the proof of usual Prikry forcing, and force $\mathsf{NS}_{\omega_1}^+$ to get a generic ultrafilter $H\ni S$. Then I applied the Ramsey-like property to get an $f$-homogeneous set in $H$. $\endgroup$
    – Hanul Jeon
    Commented Feb 21 at 0:03

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The statement implies that for any $f : [\omega_1]^2 \to 2$, there is an uncountable $S$ such that $f$ is constant on $[S]^2$. This is impossible since $\omega_1$ is not weakly compact. See Jech, Lemma 9.4 for a proof.

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  • $\begingroup$ It turned out that I asked a silly question... Thank you for your answer, though. $\endgroup$
    – Hanul Jeon
    Commented Feb 21 at 18:20

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