Hello, I have the following question (for definitions see at the end):


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 $\alpha<\kappa$ of uncountable cofinality, both $A$ and $B$ reflect at $\alpha$?


**Definitions**: (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 $\alpha$.



Disjoint stationary sets that reflect
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. 


**Definition.** The global square principle $\square$ is the
assertion that there is an assignment $\nu\mapsto C\_\nu$ for all
singular ordinals $\nu$, such that


* $C\_\nu$ is a closed subset of $\nu$, containing only singular
ordinals;
* If $\nu$ has uncountable cofinality, then $C\_\nu$ is unbounded
in $\nu$;
* the order type of $C\_\nu$ is less than $\nu$;
* and if $\mu\in C\_\nu$, then $C\_\mu=C\_\nu\cap\mu$.


(For reference, see definition 19 of [Square in Core Models](http://www.google.com/url?sa=t&rct=j&q=set%20theory%20square%20principle&source=web&cd=1&cad=rja&ved=0CDAQFjAA&url=http%3A%2F%2Fwww.math.ucla.edu%2F~asl%2Fbsl%2F0703%2F0703-001.ps&ei=KWuYUNSmGaSI0QGMt4DgDQ&usg=AFQjCNEhMwQ3gvh1pg8UsmCmsDHiM-uZzQ), by Schimmerling and Zeman, or numerous other accounts.) 


**Theorem.** 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 $\kappa$.


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 $\eta<\nu$. 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.


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
$\beta<\kappa$. 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


Since $\square$ holds in $L$, this means that ZFC+V=L proves the
affirmative answer.


(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.)


