Added to Favorites

Related Searches

Definitions

Nearby Words

In mathematics, particularly in set theory and model theory, there are at least three notions of stationary set:
## Classical notion

If $kappa$ is a cardinal of uncountable cofinality, $Csubseteqkappa$, and $C$ intersects every club in $kappa$, then $C$ is called a stationary set. If $C$ is not stationary then it is a thin set.## Jech's notion

There is also a notion of stationary subset of $[X]^lambda$, for $lambda$ a cardinal and $X$ a set such that $|X|gelambda$, where $[X]^lambda=\{Ysubset\; X:|Y|=lambda\}$. This notion is due to Thomas Jech. As before, $Ssubset[X]^lambda$ is stationary if and only if it meets every club, where a club subset of $[X]^lambda$ is a set unbounded under $subset$ and closed under union of chains of length at most $lambda$. These notions are in general different, although for $X=omega\_1$ and $lambda=aleph\_0$ they coincide in the sense that $Ssubset[omega\_1]^omega$ is stationary if and only if $Scapomega\_1$ is stationary in $omega\_1$. ## Generalized notion

There is yet a third notion, model theoretic in nature and sometimes referred to as generalized stationarity. This notion is probably due to Magidor, Foreman and Shelah and has also been used prominently by Woodin. ## References

Matthew Foreman, Stationary sets, Chang's Conjecture and partition theory, in Set Theory (The Hajnal Conference) DIMACS Ser. Discrete Math. Theoret. Comp. Sci., 58, Amer. Math. Soc. , Providence, RI. 2002 pp. 73-94
File at
## External links

In fact the intersection of a stationary set and a club set is itself stationary. This is true because if S is stationary and $C\_1\; ,\; C\_2$ are club sets we have: $S\; cap\; (C\_1\; cap\; C\_2)\; =\; (S\; cap\; C\_1)\; cap\; C\_2$. Now $C\_1\; cap\; C\_2$ is a club set as it is the intersection of two club sets. So $S\; cap\; (C\_1\; cap\; C\_2)$ is non empty. But then $(S\; cap\; C\_1)$ must be stationary as $C\_2$ is arbitrary.

See also: Fodor's lemma

The restriction to uncountable cofinality is in order to avoid trivialities: Suppose $kappa$ has countable cofinality. Then $Ssubsetkappa$ is stationary in $kappa$ if and only if $kappasetminus\; S$ is bounded in $kappa$. In particular, if the cofinality of $kappa$ is $omega=aleph\_0$, then any two stationary subsets of $kappa$ have stationary intersection.

This is no longer the case if the cofinality of $kappa$ is uncountable. In fact, suppose $kappa$ is regular and $Ssubsetkappa$ is stationary. Then $S$ can be partitioned into $kappa$ many disjoint stationary sets. This result is due to Solovay. If $kappa$ is a successor cardinal, this result is due to Ulam and is easily shown by means of what is called an Ulam matrix.

The appropriate version of Fodor's lemma also holds for this notion.

Now let $X$ be a nonempty set. A set $Csubset\{mathcal\; P\}(X)$ is club (closed and unbounded) if and only if there is a function $F:[X]^\{\}to\; x\; math>\; such\; that$ C=\{z:F[[z]^\{\}]subset\; z\}\; math>.\; Here,$ [y]^\{\}\; math>\; is\; the\; collection\; of\; finite\; subsets\; of$ y$.$\}>$$

$Ssubset\{mathcal\; P\}(X)$ is stationary in $\{mathcal\; P\}(X)$ if and only if it meets every club subset of $\{mathcal\; P\}(X)$.

To see the connection with model theory, notice that if $M$ is a structure with universe $X$ in a countable language and $F$ is a Skolem function for $M$, then a stationary $S$ must contain an elementary substructure of $M$. In fact, $Ssubset\{mathcal\; P\}(X)$ is stationary if and only if for any such structure $M$ there is an elementary substructure of $M$ that belongs to $S$.

Wikipedia, the free encyclopedia © 2001-2006 Wikipedia contributors (Disclaimer)

This article is licensed under the GNU Free Documentation License.

Last updated on Monday February 11, 2008 at 13:19:45 PST (GMT -0800)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

This article is licensed under the GNU Free Documentation License.

Last updated on Monday February 11, 2008 at 13:19:45 PST (GMT -0800)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

Copyright © 2014 Dictionary.com, LLC. All rights reserved.