# weakly normal filters

Kanamori (Ultrafilters over Uncountable Cardinals) in his Phd Thesis defines a filter $$\mathcal F$$ as weakly normal whenever every function $$f$$ such that $$\{\xi<\kappa\mid f(\xi)<\xi\}\in\mathcal F$$ then there is some $$\eta<\kappa$$ such that $$X=\{\xi<\kappa \mid f(\xi)<\eta\}$$ has positive measure (i.e. $$X$$ has non-empty intersection with every element of $$\mathcal F$$).

Then he says that weak normality is equivalent to say that every (filter) extension of $$\mathcal F$$ is weakly normal. But he states it as obvious. I can't find a justification of this.

• Are you sure you got the right definition of weakly normal? In his paper "weakly normal filters and irregular ultrafilters" Kanamori defines weakly normal to mean: If f is regressive on $\mathcal F$ then $f^{-1}[\eta]\in\mathcal F$ for some $\eta<\kappa$ (i.e. this set is in $\mathcal F$ not just of positive measure). Then it is quite clear that what you state is true. – Andreas Lietz Jan 19 at 9:59
• @user457161 can you tell me where exactly he says that? – edgar alonso Jan 19 at 14:56
• @user457161i've read this paper and he never states $f^{-1}[\eta]\in\mathcal F$ – edgar alonso Jan 19 at 15:02
• In Definition 1.1 he defines in $(v)$ that $\mathcal F$ is weakly normal iff every function that is regressive on a set in $\mathcal F$ is not unbounded (mod $\mathcal F$). In $(i)$, he states that $f:\kappa\rightarrow\kappa$ is unbounded iff for all $\eta<\kappa$, $\{\alpha<\kappa\mid f(\alpha)>\eta\}$ has positive $\mathcal F$-measure. In other words, $f$ is not unbounded, if there is $\eta$ so that $\{\alpha<\kappa\mid f(\alpha)\leq\eta\}\in \mathcal F$. Putting these two together, we get the definition I mentioned above. – Andreas Lietz Jan 19 at 18:51
• If $f$ is nounded then there is even come $\eta$ so that $\{\xi<\kappa\vert f(\xi)\leq\eta\}\in\mathcal F$ and not just of positive measure. If $f$ is unbounded, this means that $\forall\eta<\kappa\ \{\xi<\kappa\vert f(\xi)>\eta\}$ is of positive $\mathcal F$-measure. Now for $f$ to not be unboudnded, there is some $\eta$ so that $A=\{\xi<\kappa\vert f(\xi)>\eta\}$ is not of positive $\mathcal F$ measure. Thus there is $X\in\mathcal F$ with $X\cap A=\emptyset$. Hence $X\subseteq\{\xi<\kappa\vert f(\xi)\leq\eta\}\in\mathcal F$. – Andreas Lietz Jan 24 at 12:37