Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\kappa$ be an infinite cardinal. We know that the notion of being closed and unbounded (club) in $\kappa$ is strictly stronger than being stationary in $\kappa$.

My question: Do you know any notion that is strictly between being a club and being a stationary set?

share|improve this question

4 Answers 4

up vote 12 down vote accepted

Here are a few intermediate notions:

  • "Being in the club filter" is strictly stronger than stationary and strictly weaker than club.

  • More generally, "being in $F$", for some fixed normal filter $F$ on $\kappa$. By normality, $F$ contains the club filter and so all such sets are stationary, but needn't be club.

  • "Being $\omega$-closed" (that is, containing all limits of countable sequences) and unbounded. Every such set is stationary, but when $\kappa\gt\omega_1$, there are $\omega$-clubs that are not club, such as the set of ordinals with cofinality $\omega$.

  • Similarly, "being $\delta$-club" or "being ${\lt}\delta$-club" for some specific $\delta\lt\kappa$ is intermediate.

  • More generally, "agreeing with a club on $E$" for some fixed stationary set $E$. That is, $B$ agrees with a club on $E$ if there is some club $C$ such that $C\cap E=B\cap E$.

In the large cardinal context, there are several more:

  • When $\kappa$ is a measurable cardinal, then "having normal measure one," meaning that the set is in some normal measure on $\kappa$, is an example. All such sets are stationary, but not all are club or even in the club filter, and if there is more than one normal measure, these sets do not even form a filter.

  • Similarly, when $\kappa$ is even stronger, say, supercompact, then "having normal measure one with respect to a normal measure induced by a $\kappa^{++}$-supercompactness ultrafilter (or whatever, as you like). That is, $\kappa\in j(A)$ for an embedding $j:V\to M$ of the desired strong type.

  • When $\kappa$ is a weaker large cardinal, say, just weakly compact, then one can say "having weak compactness measure one." A set $A\subset\kappa$ has this property when for every $\kappa$-model $M$ there is a weak compactness embedding $j:M\to N$ with $\kappa\in j(A)$.

share|improve this answer
Thank you. You provided plenty of examples. –  Ioannis Souldatos Jun 28 '13 at 14:42
If other people have more to add, I would be interested in seeing more. –  Ioannis Souldatos Jun 28 '13 at 14:43

Along the lines of Joel's last example, Baumgartner has studied ideals associated with vrious large-cardinal properties. See in particular his papers "Ineffability properties of cardinals, I and II".

There's also an ideal associated to Jensen's diamond principle, giving rise to the following notion of largeness. Call a subset $X$ of $\omega_1$ a diamond-set if there is a sequence of sets $\langle A_\alpha:\alpha \in X\rangle$ such that each $A_\alpha\subseteq\alpha$ and, for every $Y\subseteq\omega_1$, the set $\{\alpha\in X:A_\alpha=Y\cap\alpha\}$ is stationary. Then all diamond-sets are stationary, and, if diamond holds, then all club sets are diamond-sets.

share|improve this answer
Similarly, there is the Vopenka filter in the context of Vopenka cardinals. –  Joel David Hamkins Jun 28 '13 at 16:18

A subset of $A$ of $\kappa$ is said to be $\alpha$-fat, if for every club $C$ in $\kappa$, the set $A\cap C$ contains a closed copy of the ordinal $\alpha$.

For $\kappa=\omega_1$, a set $S$ is stationary iff it is $\alpha$-fat for every $\alpha<\omega_1$ (see Friedman's PAMS paper from 1974).

For $\kappa=\omega_2$, under $V=L$, every stationary $S\subseteq\omega_2$ consisting of ordinals of countable cofinality, contains a stationary subset $S'\subseteq S$ which is not $\omega_1$-fat. On the other extreme, Martin Maximum implies that every stationary subset of $\{\alpha<\omega_2\mid cf(\alpha)=\omega\}$ is $\omega_1$-fat.

share|improve this answer
I hope you'll be back soon. Maybe we can actually get the seminar going again... –  Asaf Karagila Jun 28 '13 at 22:25

Really what you're seeing in most of the above examples is that whenever you have a proper ideal $I$ extending the non-stationary ideal on $\kappa$, then ``not being in $I$'' gives you a notion intermediate between stationary and club. So realy we might ask instead for examples of natural proper ideals $I$ on $\kappa$ that extend the non-stationary ideal.

One family of such ideals I haven't seen mentioned yet arises from the theory of club-guessing. Concentrating on one particular case, suppose $\kappa$ is a regular cardinal greater than $\omega_1$, and let $S$ be the stationary subset of $\kappa$ consisting of ordinals of countable cofinality.

There is sequence $\langle C_\delta:\delta\in S\rangle$ such that each $C_\delta$ is club in $\delta$ of order-type $\omega$, and for every closed unbounded $E\subseteq\kappa$ there are stationarily many $\delta\in S$ with $C_\delta\setminus E$ finite. (So $C_\delta$ is almost contained in $E$ for stationarily many $\delta\in S$.)

There is a natural way to get a normal filter out of the above set-up: For $E\subseteq\kappa$ club, let $S_E=\{\delta\in S: C_\delta\subseteq^* E\}$. These sets generate a normal filter $F$ concentrating on the stationary set $S$, and the condition ``positive with respect to $F$'' gives you a nice intermediate notion.

Now look at the collection of $A\subseteq\kappa$ with the property that for $F$-almost all $\delta\in S$, $A\cap C_\delta$ is finite. (These are sets that ``run away'' from the sequence $\langle C_\delta:\delta\in S\rangle$.)

This collection generates a proper ideal $I$ on $\kappa$, and any $I$-positive set is also stationary: if $A\notin I$ and $E\subseteq\kappa$ is club, we can find a $\delta$ for which $C_\delta\subseteq^* E$ and $A\cap C_\delta$ is infinite, so $A\cap E\neq\emptyset$.

share|improve this answer
+1. Some of the properties mentioned also have the form "not being in one of the $I$'s" for a family of ideals. –  Joel David Hamkins Jun 30 '13 at 23:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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