## 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 ${\kappa.}$

## 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. ${\sf QED}$