4
$\begingroup$

In what follows, $\|\Phi\|_I$ means the Galvin-Hajnal rank of a function $\Phi:\kappa \to Ord$ with respect to a $\omega_1$-complete ideal $I$ of $\kappa$.

Lemma 2.2.5 of Introduction to Cardinal Arithmetic by Holz, Steffens and Weitz says:

Assume that $\kappa > \omega$ is a regular cardinal, $\Phi \in \,^\kappa Ord$, and $I$ is a $\kappa$-complete normal ideal on $\kappa$ with $\bigcup I=\kappa$. Then $\|\Phi\|_I \leq \sigma \Leftrightarrow \{\xi < \kappa : \Phi(\xi) \leq \sigma\} \notin I$ for every ordinal $\sigma < \kappa$.

The assumption on $\kappa$-completeness cannot be dropped. But can you drop one or more of the following assumptions:

  1. $\kappa$ is regular,
  2. $\bigcup I=\kappa$ and
  3. $I$ is normal?
  • 0
    If you follow the proof closely you will see that you cannot drop the normality (i.e. the ideal is closed under diagonal unions) and this in turn implies that $\bigcup I=\kappa$. The regularity needs to stay to ensure that the ideal is normal, otherwise we need to handle cofinality issues.2011-06-17
  • 0
    It is true that in the $\Leftarrow$ direction the proof uses normality in an essential way, but maybe another proof doesn't need it... I wonder if there is an example of such a $\kappa$-complete ideal that the equivalence of the lemma does not hold.2011-06-17
  • 0
    1. If $\kappa$ is singular, but we still require that $I$ is $\kappa$-complete and $\bigcup I = \kappa$, then $\kappa \in I$ and so the problem becomes uninteresting.2011-06-18
  • 0
    2. We can drop $\bigcup I = \kappa$ iff the order type of $\bigcup I$ is $\kappa$ and you replace $\xi < \kappa$ with $\xi \in \bigcup I$ in the problem and in the definition of $f <_I g$.2011-06-18
  • 0
    @Amit Kumar Gupta: Regarding your second comment, $I=\{\emptyset\}$ is a $\kappa$-complete normal ideal on $\kappa$ and in this case $f <_I g$ means $f < g$ everywhere and, if I'm not mistaken, $|f|_I=\min_{\alpha<\kappa} f(\alpha)$. Then the equivalence of the lemma holds.2011-06-18
  • 0
    @LostInMath, yes, in general if $\kappa \supsetneq \bigcup I = X \in I$, then $f\ <_I\ g$ iff $g$ dominates $f$ on all of $\kappa - X$ and $|f|_I$ is the minimum value $f$ attains on $\kappa - X$.2011-06-18

0 Answers 0