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It is easily shown that, for any uncountable infinite cardinal $\kappa$, $\square_\kappa$ implies that for any stationary $S\subseteq \kappa^+$, there exists a stationary $T\subseteq S$ such that $T$ does not reflect at (i.e. is not stationary in) any $\alpha<\kappa$ of uncountable cofinality. The standard proof does not go through, however, when $\square_\kappa$ is replaced by the weaker notion of $\square(\kappa^+)$. Is $\square(\kappa^+)$ compatible with stationary reflection? More precisely, if $\kappa$ is an uncountable infinite cardinal, is $\square(\kappa^+)$ consistent with the statement "every stationary $S\subseteq \kappa^+$ consisting of ordinals of cofinality $<\kappa$ reflects at some $\alpha<\kappa^+$"?

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For reference, Andres Caicedo has a nice post with some discussion of the difference between the principles.… – Joel David Hamkins Aug 27 '12 at 21:13
up vote 6 down vote accepted

The answer is yes (at least for $\kappa = \omega_1$): Shelah and Harrington showed that you can force that every stationary subset of $S_{\omega}^{\omega_2}$ reflects starting with a Mahlo cardinal. See Theorem A in Some exact equiconsistency results is set theory.

Since $\neg \square(\omega_2)$ implies that $\omega_2$ is weakly compact in $L$, if we start with a Mahlo cardinal $\kappa$ which is not weakly compact in $L$ and collapse it to $\omega_2$ using the forcing of Shelah-Harrington we'll have $\square(\omega_2)$ in the generic extension.

I don't know if this is true for $\kappa > \omega_1$ as well.

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Yes, a Harrington-Shelah style forcing construction works for every regular, uncountable $\kappa$. In fact, it can be shown that, in the generic extension, we have a $\square(\kappa^+)$ sequence whose clubs avoid a stationary subset of $S^{\kappa^+}_\kappa$. I'm not sure about the situation for singular $\kappa$, though. – Chris Lambie-Hanson Oct 28 '13 at 20:26

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