Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\alpha<\kappa^+$"?

share|improve this question
For reference, Andres Caicedo has a nice post with some discussion of the difference between the principles. andrescaicedo.wordpress.com/2012/05/21/… –  Joel David Hamkins Aug 27 '12 at 21:13
add comment

1 Answer

up vote 5 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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.