Disjoint stationary sets that reflect - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T11:36:28Z http://mathoverflow.net/feeds/question/111327 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111327/disjoint-stationary-sets-that-reflect Disjoint stationary sets that reflect Ioannis Souldatos 2012-11-03T01:08:40Z 2012-11-06T21:19:38Z <p>Hello, I have the following question (for definitions see at the end):</p> <p>Let $\kappa$ be an uncountable regular cardinal. Can we prove in ZFC that there exist two disjoint stationary sets $A$, $B$ such that for every limit ordinal $\alpha&lt;\kappa$ of uncountable cofinality, both $A$ and $B$ reflect at $\alpha$?</p> <p><strong>Definitions</strong>: (1) $A$ is a stationary set on $\kappa$, if $A\subset\kappa$ and $A$ intersects every closed and unbounded set in $\kappa$. (2) A set is closed if it contains its limit points. (3) A stationary set $A\subset\kappa$ reflects at $\alpha$ if $A\cap\alpha$ is stationary on $\alpha$.</p> http://mathoverflow.net/questions/111327/disjoint-stationary-sets-that-reflect/111584#111584 Answer by Eran for Disjoint stationary sets that reflect Eran 2012-11-05T21:37:01Z 2012-11-05T21:56:38Z <p>Let's take an example - $\kappa = \omega_2$. The set $D$ of all ordinals less that $\omega_2$ with cofinality $\omega_1$ is stationary in $\omega_2$. Split it to two stationary sets A and B (using Solovay's theorem?). Now take for each ordinal in these sets an $\omega_1$ cofinal series, and split it (based on even and odd indices), between the two sets. I believe A and B now provide the requirement.</p> http://mathoverflow.net/questions/111327/disjoint-stationary-sets-that-reflect/111601#111601 Answer by Joel David Hamkins for Disjoint stationary sets that reflect Joel David Hamkins 2012-11-06T01:46:25Z 2012-11-06T02:50:45Z <p>A general affirmative answer is possible if one assumes the global square principle, which holds in $L$ and in many other canonical models. Indeed, the failure of $\square$ is a strong hypothesis. </p> <p><b>Definition.</b> The global square principle $\square$ is the assertion that there is an assignment $\nu\mapsto C_\nu$ for all singular ordinals $\nu$, such that</p> <ul> <li>$C_\nu$ is a closed subset of $\nu$, containing only singular ordinals;</li> <li>If $\nu$ has uncountable cofinality, then $C_\nu$ is unbounded in $\nu$;</li> <li>the order type of $C_\nu$ is less than $\nu$;</li> <li>and if $\mu\in C_\nu$, then $C_\mu=C_\nu\cap\mu$.</li> </ul> <p>(For reference, see definition 19 of <a href="http://www.google.com/url?sa=t&amp;rct=j&amp;q=set%20theory%20square%20principle&amp;source=web&amp;cd=1&amp;cad=rja&amp;ved=0CDAQFjAA&amp;url=http%3A%2F%2Fwww.math.ucla.edu%2F~asl%2Fbsl%2F0703%2F0703-001.ps&amp;ei=KWuYUNSmGaSI0QGMt4DgDQ&amp;usg=AFQjCNEhMwQ3gvh1pg8UsmCmsDHiM-uZzQ" rel="nofollow">Square in Core Models</a>, by Schimmerling and Zeman, or numerous other accounts.) </p> <p><b>Theorem.</b> If the global square $\square$ principle holds, then the answer to the question is yes, every $\kappa$ has such a partition. Indeed, under $\square$ there is a coherent global partition of the class of singular ordinals into $A\sqcup B$, such that for every $\kappa$ of uncountable cofinality, both $A\cap\kappa$ and $B\cap\kappa$ are stationary in $\kappa$.</p> <p>Proof. Fix the $\square$ sequence $C_\nu$. First, define $A$ and $B$ up to $\omega_1$ to be any partition of the singular countable ordinals into stationary sets. Suppose now that $A$ and $B$ are defined up to $\nu$, a singular limit ordinal. Consider $C_\nu$, which has some order type $\eta&lt;\nu$. If $\eta\in A$, then put $\nu\in A$, otherwise, put $\nu\in B$. Continue by transfinite recursion. Note that $A$ and $B$ partition the singular ordinals.</p> <p>Suppose that $\kappa$ has uncountable cofinality. If $\kappa=\omega_1$, then $A\cap\kappa$ and $B\cap\kappa$ are the stationary sets that we used to start the construction. More generally, if $\kappa\gt\omega_1$ but has cofinality $\omega_1$, then $\kappa$ is singular and so $C_\kappa$ is a club of some type $\beta&lt;\kappa$. Further, $A$ and $B$ when restricted to $C_\kappa$ are copies of $A\cap\beta$ and $B\cap\beta$, which by induction are each stationary. So $A\cap\kappa$ and $B\cap\kappa$ are stationary. Finally, we have the case that $\kappa$ has cofinality larger than $\omega_1$. Fix any club $C\subset\kappa$. Thus, there is some singular $\eta\in C$ with uncountable cofinality. So $C_\eta\cap C$ is club in $\eta$ and thus meets both $A$ and $B$, and so $C$ meets both $A$ and $B$, as desired. QED</p> <p>Since $\square$ holds in $L$, this means that ZFC+V=L proves the affirmative answer.</p> <p>(Click on the edit history to see my original answer, which handles just the case for $\kappa=\omega_2$, assuming $\square_{\omega_1}$. The idea here follows something like the idea of Eran's construction, but seems to require $\square$ in order to avoid the incoherence issue mentioned by Andreas in the comments.)</p> http://mathoverflow.net/questions/111327/disjoint-stationary-sets-that-reflect/111679#111679 Answer by Todd Eisworth for Disjoint stationary sets that reflect Todd Eisworth 2012-11-06T20:24:01Z 2012-11-06T21:19:38Z <p>In the presence of large cardinals, one can (or rather Shelah can...) force the answer to be "NO" in a very strong sense. The place to look is Section 7 of Chapter X of Proper and Improper Forcing.</p> <p>In particular, Theorem 7.4 shows that assuming the consistency of 2 supercompact cardinals, one can force that for any regular $\kappa>\omega_1$, any stationary subset of $S^\kappa_{\aleph_0}$ contains a closed copy of $\omega_1$.</p> <p>This implies the answer to your question is no by the following argument: </p> <p>Step 1: If $\kappa>\aleph_1$ is regular and $A$ reflects at all uncountable limit ordinals below $\kappa$, then so does $A\cap S^\kappa_{\aleph_0}$ (where $S^\kappa_\tau$ is the set of ordinals less than $\kappa$ of cofinality $\tau$).</p> <p>Proof: Let $A_0= A\cap S^\kappa_0$, and let $A_1= A\setminus A_0$. $A_1$ cannot reflect at ordinals of cofinality $\omega_1$, and so it must be the case that $A_0$ reflects at all ordinals of cofinality $\omega_1$. But then $A_0$ also reflects at any place where $S^\kappa_{\aleph_1}$ reflects as well, and so $A_0$ reflects at all ordinals of uncountable cofinality below $\kappa$.</p> <p>Step 2:<br> Assume we are in a model like that obtained by Shelah. If $\kappa$ is a regular cardinal greater than $\aleph_1$ and $A$ is a stationary subset of $S^\kappa_{\aleph_0}$. We know $A$ contains a closed copy $C$ of $\omega_1$, and if we set $\delta=\sup(C)$ then $\delta$ is an ordinal of cofinality $\omega_1$ where $A$ reflects but $\kappa\setminus A$ does not. In particular, no stationary subset disjoint to $A$ can reflect at $\delta$, hence there is no way to get your "$B"$.</p> <p>Edit:</p> <p>A "no" answer to your question at $\omega_2$ is equiconsistent with the existence of a Mahlo cardinal.</p> <p>As Joel mentioned in (an earlier version of) his answer, one can build $A$ and $B$ in $\omega_2$ from a $\square_{\omega_1}$-sequence. The failure of $\square_{\omega_1}$ implies that $\aleph_2$ is Mahlo in $L$ (Credited to Jensen on page 453 of Jech's "Set Theory"; I don't know a better reference.)</p> <p>On the other hand, Theorem 7.1 in Chapter XI (page 576) of Proper and Improper forcing tells us that from a Mahlo cardinal, we can force ZFC+GCH + "every stationary subset of $S^{\omega_2}_{\omega}$ contains a closed copy of $\omega_1$, which we argued above gives a "No" answer.</p> <p>Note that what Shelah is really showing is the consistency of the following statement:</p> <p>"If $S$ is a stationary subset of $S^{\omega_2}_{\omega}$ that reflects at every member of $S^{\omega_2}_{\omega_1}$, then $S^{\omega_2}_{\omega}\setminus S$ is non-stationary,"</p> <p>while the original question is equivalent to asking of $S^\kappa_\omega$ can be partitioned into two disjoint stationary sets, each of which reflects at every ordinal in $S^\kappa_{\omega_1}$.</p>