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Let $\kappa$ be a cardinal (of uncountable cofinality). A subset $S \subseteq \kappa$ is called stationary if it intersects every club, i.e. closed unbounded subset of $\kappa$. Now my question is basically just: Why do we care about stationary sets? I know some statements, which are independent from ZFC, for example the diamond principle, which involve them, but what was or is the motivation for studying them? Please don't just give an overview of the results, because they can be found in every textbook on cardinals.

Also, what is your intuition for stationary sets? I think clubs are very easy to grasp; they just contain big enough ordinals. Perhaps stationary sets follow the same idea, but in a "second order"; they contain enough big enough ordinals?

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Some intuition might be given by the following informal analogy with measure theory: if we have a measure space of measure 1, then club sets are analogous to subsets of measure 1, while stationary sets are analogous to sets of positive measure.

In other words, club sets contain "almost all" ordinals, while stationary sets contain "a positive proportion" of them.

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    $\begingroup$ It seems more than just an informal analogy, since there is an actual $\sigma$-additive two-valued measure here on the $\sigma$-algebra of sets that either contain or omit a club. $\endgroup$ – Joel David Hamkins Sep 2 '10 at 15:37
  • $\begingroup$ Maybe, but one should not look too closely at the phrases "positive measure" or "positive proportion". $\endgroup$ – Richard Borcherds Sep 2 '10 at 23:55
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    $\begingroup$ It's not maybe: define a set has measure $1$ if it contains a club, and measure $0$ if it omits a club. This is a $\sigma$-additive measure on a $\sigma$-algebra of subsets of $\omega_1$, and the non-measure zero sets are precisely the stationary sets. $\endgroup$ – Joel David Hamkins Dec 28 '16 at 17:08
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One answer to your question about intuition is simply that stationary sets arise very naturally once you begin to think of the natural measure surrounding club sets. The stationary sets are simply those that have positive outer measure with respect to the club filter. So if you care about club sets being large, then the concept of stationary sets arises naturally.

More specifically, if you consider the collection of sets that either contain or omit a club, then you have a natural two-valued measure (measure 1 means containing a club, measure 0 means omitting a club). The stationary sets are precisely the sets that do not have measure 0, and this is the same as having outer measure 1.

Many uses of the club sets rely on the fact that they can be thought of as the large subsets of $\kappa$, in these sense that they have measure 1 with respect to this measure. However, many of these uses generalize from club to stationary because it is sufficient for the application that the set is larger merely in the sense that it does not have measure $0$, rather than actually have measure $1$.

By the way, I'm not so sure I share your intuition that the club sets are those that "contain big enough ordinals". On the one hand, if $C$ is the set of limit ordinals below $\kappa$, then it is club, but if I add one (or any fixed non-zero ordinal) to every ordinal in $C$, making $D=\{\lambda+1\ |\ \lambda\in C\}$, then it would seem to have just as big ordinals or bigger, but $D$ is not club.

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    $\begingroup$ Another reason to care about stationary sets: in the case of $\omega_1$, the stationary sets are precisely the sets that could have a club subset in a forcing extension. That is, we can force to shoot a club through any given stationary set, by forcing that preserves $\omega_1$. $\endgroup$ – Joel David Hamkins Sep 2 '10 at 15:28
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For purposes of intuition, I've found the following reformulation of the definition useful. A subset $S$ of $\kappa$ is stationary if and only if, for any countably many finitary operations on $\kappa$, say $f_n:\kappa^{r_n}\to\kappa$, there is an $\alpha\in S$ closed under all the $f_n$'s. In the terminology of universal algebra, this says that every algebra with underlying set $\kappa$ and with countable type has a subalgebra that is a member of $S$. By using Skolem functions, one can also rephrase the definition in model-theoretic terms: Any structure with universe $\kappa$ for a countable language has an elementary submodel whose universe is a member of $S$. (That bring us into the general neighborhood of the somewhat deeper comments in Andres Caicedo's answer.)

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  • $\begingroup$ Hi Andreas. A nice advantage of this approach is that it may help understand the combinatorics behind the general notion of stationarity without too much additional effort. $\endgroup$ – Andrés E. Caicedo Nov 10 '10 at 0:01
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The way I tend to think of/use them is the following: Any ``reasonable'' property (of ordinals less than a cardinal $\kappa$) holds on a club, so for any such property a stationary set is guaranteed to have points satisfying that property.

This is very useful when using ``model theoretic'' arguments. Typically, you want to prove some property of $V$ by passing to an elementary substructure of a sufficiently large initial segment of the universe. [This in itself is a really useful technique that is worth investing time in.] By using a chain of models argument, you can ensure that certain ordinals associated to your structure are in a given stationary set. This typically guides the combinatorics of the argument, and is particularly handy in forcing arguments.

For example, if $S$ is stationary in $\omega_1$ and $C$ is club, for any countable elementary substructure $M$ of some large segment of $V$ such that $C\in M$, we automatically have that $M\cap\omega_1$ is an ordinal that belongs to $C$. So in a chain of such models, stationarily many have their $\omega_1$ in $S$. This is useful in showing that certain forcing notions defined in terms of $S$ preserve $\omega_1$, which is a very desirable property to have.

An example of a different kind that I like: Boban Velickovic proved years ago that if MM holds and $M$ is an inner model with the same $\omega_2$ then $M$ contains all the reals. A key feature of his argument is a game on ordinals. The game depends on a parameter $\alpha\in\omega_1$; call it $G_\alpha$. These games are designed in a way that ensures they are determined. One can then prove that the player we care about wins $G_\alpha$ for stationarily many values of $\alpha$. This ensures that we can take a countable elementary $N$ with $N\cap\omega_1=\alpha$ being such that the good player wins $G_\alpha$. The winning strategy for that player is then used to guide a coding construction, that eventually ensures that we can capture any real of $V$ in $M$. This is in his paper "Forcing axioms and stationary sets", Advances in Mathematics, vol. 94 (2), (1992), pp. 256-284.

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You specifically asked not to give an overview of the results, but I will mention one important result, Fodor's lemma (pressing down lemma) which is important (and provable in ZFC).

Let $\kappa$ be an uncountable regular cardinal and $S\subseteq\kappa$. A function $f:S\to\kappa$ is regressive (pressing down) if for all $\alpha\in S$, $f(\alpha)<\alpha$.
Now Fodor's lemma says that every regressive function on a stationary set is constant on a (possibly smaller) stationary set.

This for example implies that if you partition a stationary set into few pieces ($<\kappa$), then one of the pieces is again stationary. This is a "closure property" that the stationary sets have, but the closed unbounded sets don't.
So, in some sense stationarity is a largeness property that is better than being closed unbounded. Another (related) example of this is that the diamond principle that you mention is consistent, but if "stationary" is replaced by "closed unbounded", then the principle becomes wrong (this can be rectified by changing another part of the definition of the principle, though).

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This answer overlaps with other answers, but I think it’s worth mentioning separately because I find, as a general principle, that one good way to motivate a concept is to see how it can be used to solve problems that ostensibly don’t involve the concept. In a short paper, A characterization of free abelian groups, J. Steprans proved that an abelian group is free if and only if it has a discrete norm. A norm is a nonnegative real-valued function $\nu$ such that

  • $\nu(g) = 0$ if and only if $g$ is the identity,
  • $\nu(g+h) \le \nu(g)+\nu(h)$, and
  • $\nu(mg) = |m|\nu(g)$ if $m\in\mathbb{Z}$.

A norm $\nu$ is discrete if there exists $\rho>0$ such that $\nu(g) > \rho$ whenever $g$ is not the identity. Steprans’s proof assumes toward a contradiction that $G$ is an abelian group $G$ that is not free but that has a discrete norm $\nu$. The cardinality $\kappa$ of $G$ may be assumed to be minimal; known results implied that $\kappa$ would have to be uncountable. A key part of the proof is to argue that $G$ has “arbitrarily large approximations” that are not free; stationary sets come into play here. More precisely, by Shelah’s compactness of singular cardinals, $\kappa$ is regular, and $G$ is an increasing union of subgroups $(G_\alpha : \alpha <\kappa)$, and there exists a stationary set $S\subseteq \kappa$ and a function $f:S\to\kappa$ such that $G_{f(\alpha)}/G_\alpha$ is not free whenever $\alpha\in S$. Then one uses the properties of stationary sets and discrete norms to deduce the existence of a smaller abelian group $G$ that is not free and that has a discrete norm, yielding a contradiction by “infinite descent.”

I learned about this example from an answer to another MO question about using set theory to prove results from other areas of mathematics.

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  • $\begingroup$ I think you linked the wrong answer. :-) $\endgroup$ – Asaf Karagila Oct 27 '20 at 13:42
  • $\begingroup$ Now that I see your answer, I agree! But as it happens, I saw Golshani's answer first and went off and read Roitman's article before looking at the other answers. $\endgroup$ – Timothy Chow Oct 27 '20 at 13:46
  • $\begingroup$ I did not follow the link to Roitman's article! $\endgroup$ – Asaf Karagila Oct 27 '20 at 13:47

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