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 $\alpha$.


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. $\endgroup$ – Ioannis Souldatos Nov 7 '12 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. $\endgroup$ – Joel David Hamkins Nov 6 '12 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... $\endgroup$ – Todd Eisworth Nov 6 '12 at 20:52
  • $\begingroup$ No idea about the case where the regular cardinal is successor of singular though! $\endgroup$ – Todd Eisworth Nov 6 '12 at 20:52
  • $\begingroup$ Very interesting! It seems there may be a very close connection in general with $\square_\kappa$... $\endgroup$ – Joel David Hamkins Nov 6 '12 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. $\endgroup$ – Ioannis Souldatos Nov 14 '12 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. $\endgroup$ – Joel David Hamkins Nov 5 '12 at 23:26
  • 3
    $\begingroup$ 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. $\endgroup$ – Andreas Blass Nov 5 '12 at 23:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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