580 -Cardinal arithmetic (8)

March 5, 2009

 

4. Large cardinals and cardinal arithmetic

 

In section 3 we saw how the powers of singular cardinals (or, at least, of singulars of uncountable cofinality) satisfy strong restrictions. Here I show that similar restrictions hold at large cardinals. There is much more than one could say about this topic, and the results I present should be seen much more like an invitation than a full story. Also, for lack of time, I won’t motivate the large cardinals we will discuss. (In the ideal world, one should probably say a few words about one’s beliefs in large cardinals, since their existence and even their consistency goes beyond what can be done in the standard system {{\sf ZFC}.} I’ll however take their existence for granted, and proceed from there.)

 

1. Measurable cardinals

 

Definition 1 {\kappa} is a measurable cardinal iff {\kappa>\omega} and there is a nonprincipal {\kappa}-complete ultrafilter over {\kappa.}

 

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580 -Cardinal arithmetic (5)

February 13, 2009

At the end of last lecture we defined club sets and showed that the diagonal intersection of club subsets of a regular cardinal is club.

Definition 10. Let \alpha be a limit ordinal of uncountable cofinality. The set S\subseteq\alpha is stationary in \alpha iff S\cap C\ne\emptyset for all club sets C\subseteq\alpha. 

For example, let \lambda be a regular cardinal strictly smaller than {\rm cf}(\alpha). Then S^\alpha_\lambda:=\{\beta<\alpha : {\rm cf}(\beta)=\lambda\} is a stationary set, since it contains the \lambda-th member of the increasing enumeration of any club in \alpha. This shows that whenever {\rm cf}(\alpha)>\omega_1, there are disjoint stationary subsets of \alpha. Below, we show a stronger result. The notion of stationarity is central to most of set theoretic combinatorics. 

Fact 11. Let S be stationary in \alpha.

  1. S is unbounded in \alpha.
  2. Let C be club in \alpha. Then S\cap C is stationary. 

Proof. 1. S must meet \kappa\setminus\alpha for all \alpha and is therefore unbounded.

2. Given any club sets C and D, (S\cap C)\cap D=S\cap(C\cap D)\ne\emptyset, and it follows that S\cap C is stationary. {\sf QED}

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