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If $\kappa>\omega_1$ is a regular cardinal, $FP_\kappa$ is the assertion that every stationary subset of $\kappa$ consisting of ordinals of countable cofinality has a closed subset of order type $\omega_1$.

Question: If $\kappa$ is supercompact or has a supercompact cardinal below it, is it consistent that $FP_\kappa$ fails?

For context: I'm trying to show that a particular strengthening of PFA does not imply MM. Friedman's property (which under MM holds at almost all regular cardinals) seems like a natural candidate for separating them. My understanding is that the standard way of forcing a failure of FP is to add a nonreflecting stationary set; unfortunately, this preserves neither my axiom nor supercompactness. It is not too hard to show that if $FP_\kappa$ fails at a supercompact cardinal $\kappa$ (or at some larger cardinal), it still holds after performing the iteration to force the axiom, but I'm having trouble showing that that premise is consistent.

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    $\begingroup$ Problem 72 in Friedman's "One Hundred and Two Problems in Mathematical Logic" states "If $\kappa$ is a measurable cardinal then every subset of $\kappa$ either contains or is disjoint from a closed subset of $\kappa$ ordertype $\omega_1$". So if a solution exists it might shed light on your problem or the problem has been open for 50 years. $\endgroup$
    – Hannes Jakob
    Commented Jan 20 at 14:06

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This is more of a long comment. In Harvey Friedman's original paper introducing the property (called "On closed sets of ordinals"), he mentions at the end that, if $M[G]$ is obtained by forcing with $\operatorname{Coll}(\omega,\omega_1)$, $FP_{\kappa}$ fails for any uncountable $\kappa$. Obviously this forcing preserves supercompact cardinals but i do not see why the property fails. They state to take the set $A_{\kappa}:=\{\alpha<\kappa\;|\;\operatorname{cf}^M(\alpha)=\omega\}$. This set is stationary in any $\kappa\geq\omega_2^M=\omega_1^{M[G]}$ by the $\omega_2$-cc.. However, i do not see why the set cannot contain a closed subset of ordertype $\omega_1^{M[G]}=\omega_2^M$. I thought their idea might have been that ``obviously'' the $\omega_1^M$th element of such a set must have $M$-cofinality $\omega_1^M$, but as far as i can tell this would also prevent $A_{\kappa}$ from having any closed subset of ordertype $\omega_1^M+1$, which cannot be the case because any stationary subset of $\omega_1^{M[G]}$ contains arbitrarily long countable closed subsets.

Additionally, Harvey Friedman himself lists the problem in his list of "One Hundred and Two Problems in Mathematical Logic" which was published around the same time.

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    $\begingroup$ To see why $FP_\kappa$ fails after forcing with $\mathrm{Coll}(\omega, \omega_1)$: work in $M$, and let $\dot{f}$ be a name for a continuous, increasing map from $\omega_2^M$ to $A_\kappa$. Then we can find a single condition $p$ that decides the value of $\dot{f}(\alpha)$ for cofinally many $\alpha < \omega_2^M$. But since $\dot{f}$ is forced to be continuous, $p$ actually decides the value of $\dot{f}$ on a club, so we have a closed subset of order type $\omega_2^M$ inside $A_\kappa$ in $M$ itself, which is a contradiction. $\endgroup$
    – Chris Lambie-Hanson
    Commented May 16 at 8:08
  • $\begingroup$ Thank you very much! Do you know perchance why this is still listed as one of the problems, despite other problems being updated to show their resolution? $\endgroup$
    – Hannes Jakob
    Commented May 16 at 8:18

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