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I haven't seen this mentioned in Jech or Kunen, sorry if these are basic questions but I am not understanding these points.

If $\kappa$ is regular uncountable then in ZFC every stationary subset of $\kappa$ is the disjoint union of $\kappa$ stationary sets. So this holds for every successor $\gamma^+$. we have strictly less than $2^{\kappa}$ of them but is there a way to count how many stationary and c.u.b sets we can have?

Is there a way of knowing which subsets of say $\omega_1$ are stationary? Do we know if a stationary set can contain a c.u.b set or the other way around or maybe it does not matter?

What about for a singular cardinal, for instance for $\aleph_{\omega}$, can we count the stationary subsets and the c.u.b sets? Would the GCH help in this case?

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I'm not sure I understand your question. There are $2^\kappa$ stationary subsets of $\kappa$ for any $\kappa$ of uncountable cofinality. There can be no more than $\kappa$ many pairwise disjoint such sets. On countable cofinality, the notion of stationarity makes not much sense. The other question, characterizing stationary subsets of $\omega_1$, is classic: They are the sets that contain a club in some $\omega_1$-preserving outer model, this is a result of Baumgartner, Harrington, and Kleinberg. –  Andres Caicedo Jul 4 '10 at 20:34
    
Ah It is true, I forgot that for a countable cofinality, stationarity does not make sense. –  Carlo Von Schnitzel Jul 4 '10 at 20:59

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For every infinite cardinal $\kappa$, the number of club subsets of $\kappa$ is $2^\kappa$, fully as large as it could possibly be. Since every club set is stationary, this means also there are fully $2^\kappa$ many stationary sets also.

To see that there are this many club sets, observe that there are $\kappa$ many successor ordinals below $\kappa$. For any set $A\subset\kappa$, let $A+1 = \{\alpha+1| \alpha\in A\}$' be the set of successors to elements of $A$. Let $C_A$ be the set consisting of $A+1$ together with all limit ordinals below $\kappa$. This is a club subset of $\kappa$, and it is easy to see that if $A\neq B$, then $C_A\neq C_B$. Since there are $2^\kappa$ many subsets $A$, it follows that there are $2^\kappa$ many club sets. This argument does not need GCH.

For stationary sets, you could have used Solovay's theorem directly. Since every stationary subset of $\kappa$ is a union of $\kappa$ many disjoint stationary sets, we can take any subunion corresponding to any subset of the index set, to get $2^\kappa$ many distinct stationary sets.

About your latter considerations. Every club set is stationary; the intersection of two clubs is club and the intersection of a stationary and a club is stationary, and these are usually counted among the elementary facts about club and stationary sets. We usually do not consider the concept of stationary on singular cardinals of cofinality $\omega$, because in this case, the clubs do not form a filter. When $\kappa$ has uncountable cofinality, then the club filter makes sense, and the stationary concept is robust. The right way to think about it is: club means having measure $1$ with respect to the club filter, and stationary means not having measure $0$. Thus, stationary sets have outer measure $1$ with respect to club filter, although if they are co-stationary, then they will have inner measure $0$.

One interesting issue is that whether or not a set $A$ is stationary can be affected by forcing. For example, for every stationary set $A\subset\omega_1$, there is an $\omega_1$-preserving forcing extension where it contains a club. Thus, if the complement of $A$ was stationary in the ground model, it becomes nonstationary in the extension.

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Thank you Joel for the answer. This clarifies things a lot. Indeed every c.u.b is stationary, but can we have a stationary set of $\kappa$ in which $C_A$ is contained? Maybe it is a subset of the union of the $\kappa$ many disjoint stationary sets all containing $A$? –  Carlo Von Schnitzel Jul 4 '10 at 20:40
    
alephomega, $C_A$ is a club and so every subset of $\kappa$ containing $C_A$ must be stationary. Also, "the" partition of a stationary into disjoint stationary sets is by no means unique, so it makes no sense to talk of "the union of the disjoint stationary sets containing A " –  Andres Caicedo Jul 4 '10 at 20:50
    
Andres, I see now: this is because being a measure 1 set implies being a positive set also (every c.u.b is stationary). It is good to know that fact about the partition not being unique, I had not thought of that. –  Carlo Von Schnitzel Jul 4 '10 at 21:12
    
Joel, since you mention forcing, if you collapse $\omega_2$ to $\omega_1$ do the stationary sets of $\omega_2$ become stationary sets of $\omega_1$ necessarily? –  Carlo Von Schnitzel Jul 4 '10 at 21:18
    
If you collapse $\omega_2$ to $\omega_1$, then the old $\omega_2$ remains an ordinal, but now of cardinality $\omega_1$. If it now has cofinality $\omega$, as in Namba forcing, then the concept of stationary for subsets is no longer sensible, and in particular, the old stationary subsets are no longer stationary. If it now has cofinality $\omega_1$, the only other alternative, then $\omega_1$ is preserved and the set of ordinals below $\omega_2^V$ with cofinality $\omega_1$ in $V$, which was stationary in $V$, is no longer stationary in the forcing extension. –  Joel David Hamkins Jul 6 '10 at 3:14

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