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 $\alpha$.


3 Answers 3


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, 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.)

  • $\begingroup$ @ Joel: Thank you for the answer. I will have to digest it as well as Todd's answer. $\endgroup$ Nov 7, 2012 at 18:52

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.

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 $\omega_1$.

This implies the answer to your question is no by the following argument:

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 $\tau$).

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 $\kappa$.

Step 2:
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 "$B"$.


A "no" answer to your question at $\omega_2$ is equiconsistent with the existence of a Mahlo cardinal.

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.)

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.

Note that what Shelah is really showing is the consistency of the following statement:

"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,"

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 $S^\kappa_{\omega_1}$.

  • $\begingroup$ Great! So our answers together show that the assertion is independent of ZFC, modulo large cardinals. I suppose that the next step is to inquire as to the consistency strength of violations of the property. $\endgroup$ Nov 6, 2012 at 20:37
  • $\begingroup$ I got curious and started reading some more. In the next chapter, Shelah (drastically) improves things at least in the $\omega_2$ case. Looks like the failure of the principle at $\omega_2$ is equiconsistent with the existence of a Mahlo cardinal. That's the same level as failure of square at $\omega_2$. I'll put details in an edit when I get time... $\endgroup$ Nov 6, 2012 at 20:52
  • $\begingroup$ No idea about the case where the regular cardinal is successor of singular though! $\endgroup$ Nov 6, 2012 at 20:52
  • $\begingroup$ Very interesting! It seems there may be a very close connection in general with $\square_\kappa$... $\endgroup$ Nov 6, 2012 at 21:13
  • $\begingroup$ @Todd: Thank you for the answer. So, the answer is "it is independent". Unfortunately, I can not check both answers as correct. I marked Joel's answer as correct since it came first. $\endgroup$ Nov 14, 2012 at 19:33

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.

  • 1
    $\begingroup$ If the limit ordinals are even, then all the limit elements in your cofinal sequence will go into $A$ and none into $B$, which will make $B$ nonreflecting at the ordinals of cofinality $\omega_1$ that are not limits of such ordinals. Also, not every splitting of $\text{Cof}_{\omega_1}$ into stationary $A$ and $B$ will have the property that they reflect at every point. So I don't think this answer is correct. $\endgroup$ Nov 5, 2012 at 23:26
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    $\begingroup$ There's a lot here that isn't clear to me. Are the "$\omega_1$ cofinal series" supposed to be continuous? If so, the even-indexed elements will be club in the supremum of the sequence and the odd ones nonstationary, so I don't see where you'll get stationarity. But if not, I don't see how you'll end up with reflecting stationary sets. Also, those series will intersect a lot, by Fodor's theorem, so I don't see where disjointness will come from. $\endgroup$ Nov 5, 2012 at 23:29

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