Let $\lambda\geq \omega_2$ be a regular cardinal and $S\subset[\lambda]^\omega$ be a stationary set. I'm looking for a property of $S$, say "shootable", such that there exists a forcing extension preserving $\lambda, \omega_1$ as cardinals that shoots a club into $S$. I've encountered ad hoc examples, but I'd really hope if there are more explicit descriptions of a stationary set being "shootable" and given that a canonically defined reasonable forcing to shoot a club through it. I'm flexible with any cardinal arithmetic assumptions.
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$\begingroup$ Have you seen examples that do not satisfy the definition of "fat" given below? $\endgroup$– Monroe EskewFeb 11, 2018 at 22:13
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1$\begingroup$ @MonroeEskew any non-reflecting stationary subset of $[\omega_2]^\omega$ is not fat. $\endgroup$– Jing ZhangFeb 12, 2018 at 22:31
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$\begingroup$ I mean examples of shootable sets. $\endgroup$– Monroe EskewFeb 12, 2018 at 22:58
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1$\begingroup$ @MonroeEskew Usuba's "Reflection principles for $\omega_2$ and the semistationary reflection principle" has another example, ultimately derived from some ideas of Sakai and Krueger. He shows that if there is a "nonreflecting ladder system" $\vec{d}$ for $\omega_2 \cap \text{cof}(\omega)$ and $S$ is a nonreflecting stationary set of $x \in [\omega_2]^\omega$ such that $d_{\text{sup}(x)} \subset x$, there is a way to kill stationarity of $S$ while preserving $\omega_1$. If CH holds then $\omega_2$ is also preserved. $\endgroup$– Sean CoxFeb 14, 2018 at 16:21
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$\begingroup$ Is it clear that the complement of $S$ does not satisfy the definition of "fat"? $\endgroup$– Jing ZhangFeb 14, 2018 at 19:14
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There is a notion of fat stationary subset of $P_\kappa(\lambda)$ and a natural forcing for shooting a club through it, that you can find in the thesis Fat subsets of $P_{\kappa}(\lambda)$ by Ivan Zaigralin. In particular look at chapter 2 of the thesis.
answered Feb 8, 2018 at 5:12
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$\begingroup$ I noticed that the thesis uses the Jech definition for clubs, and while this is fine for the case in question ($\kappa=\omega$), I seem to recall that this is weaker than the algebraic definition in general. $\endgroup$– Asaf Karagila ♦Feb 8, 2018 at 7:08