# Normal ultrafilters on $\mathcal{P}_{\kappa}(\lambda)$

I am looking at fine normal ultrafilters on $\mathcal{P}_{\kappa}(\lambda)$ and I have seen that there are two (seemingly) different formulations of normality, in terms of regressive (sometimes also called choice") functions.

Note that I am aware of the usual definition of normality in terms of diagonal intersections.

In what follows, fix (infinite) cardinals $\kappa<\lambda$, and assume that $\mathcal{U}$ is a $\kappa$-complete fine ultrafilter on $\mathcal{P}_{\kappa}(\lambda)$.

One formulation of normality is the one in Jech's book (Ch. 20, Definition 20.12). Namely, $\mathcal{U}$ is normal if and only if for every function $f:\mathcal{P}_{\kappa}(\lambda)\longrightarrow\lambda$ such that $f$ is regressive (i.e., $f(x)\in x$) for $\mathcal{U}$-almost all $x$, there is some $\alpha<\lambda$ such that $f(x)=\alpha$ for $\mathcal{U}$-almost all $x$. Note that this definition talks about something happening for $\mathcal{U}$-almost all $x$", i.e., on a set belonging to the ultrafilter.

The other (apparently equivalent) formulation of normality is given in Kanamori's book (Ch. 22, Exercise 22.6). There, $\mathcal{U}$ is normal if and only if for every function $f:\mathcal{P}_{\kappa}(\lambda)\longrightarrow\lambda$ such that $f$ is regressive (i.e., $f(x)\in x$) on a $\mathcal{U}$-stationary set (where $X\subseteq\mathcal{P}_{\kappa}(\lambda)$ is $\mathcal{U}$-stationary if $X\cap Z\neq\varnothing$, for all $Z\in\mathcal{U}$), there is some $\alpha<\lambda$ such that $f(x)=\alpha$ on a $\mathcal{U}$-stationary set.

My question is about the difference in these two formulations, and we are they equivalent.

In particular, I guess that something has to be said about sets belonging to the ultrafilter $\mathcal{U}$ vs. sets being $\mathcal{U}$-stationary. It is clear that every set that belongs to a filter $\mathcal{F}$ is certainly $\mathcal{F}$-stationary. On the other hand, in general, $\mathcal{F}$-stationary sets do not necessarily belong to the filter $\mathcal{F}$. For example, let $\mathcal{F}$ be the club filter on a regular uncountable cardinal $\kappa$, where one has disjoint stationary subsets of $\kappa$.

Can someone please clarify this situation?

## 1 Answer

Clearly every set in the ultrafilter is stationary, that is part of the definition of a filter to begin with; but on the other hand, if a set has non-empty intersection with all the members of an ultrafilter it has to be there as well, otherwise its complement is in the ultrafilter...

Kanamori's definition is more general, because it works for filters, e.g. the club filter is normal with respect to that definition. But it is rarely that the club filter is an ultrafilter.