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Sep 14, 2012 at 18:04 comment added Monroe Eskew Asaf, it is not equivalent. Suppose $I$ is a $\kappa^+$-saturated ideal on a regular $\kappa$. Let $f$ be a bijection between the limit ordinals and successor ordinals of $\kappa$. Then $f$ projects $I$ to a $\kappa$-complete, $\kappa^+$-saturated ideal $J$ for which the set of successor ordinals is measure one. $J$ is not normal but $P(\kappa)/J$ is a complete boolean algebra.
Sep 14, 2012 at 17:51 vote accept Asaf Karagila
Sep 14, 2012 at 17:33 comment added Asaf Karagila Ah, I see. Thanks Joseph and Sean. Is normality somehow equivalent to completeness (or some $\kappa^+$-closedness) of the quotient algebra?
Sep 14, 2012 at 17:02 comment added Sean Cox Regarding Asaf's questions: 1) I don't know, other than viewing elements of a Boolean algebra as being 0, 1, or "positive". 2) It's just a very nice property of normal ideals. For non-normal ideals on $\kappa$, sups and infs of $\kappa$-sized subsets of $\mathbb{B}_I$ do not necessarily exist, as shown by Joseph's example (his example was a non-normal ideal on $\kappa = \omega$, but more generally, for any regular $\kappa$ the ideal $J$ of bounded subsets of $\kappa$ is $\kappa$-complete, non-normal, and some $\kappa$-sized subsets of $\mathbb{B}_{J}$ fail to have sups).
Sep 14, 2012 at 16:06 comment added Joseph Van Name I do not think there is a nice notion of $inf[X_{a}]$ in arbitrary ideals $I$ since the $inf[X_{\alpha}]$ generally does not exist. For instance, in $P(\omega)/fin$ where $fin$ is the ideal consisting of finite sets, then the least upper bound of a countable sequence of elements almost never exists since there is no countable strictly increasing sequence of elements $(x_{n})$ in $P(\omega)/fin$ with a least upper bound.
Sep 14, 2012 at 14:57 comment added Asaf Karagila Hm, that is quite a wonderful insight. I have two followup questions now: (1) Is there yet a measure theoretic notion corresponding to that? (2) If we take an arbitrary ideal $I$, is there a reasonable way to describe the operation $\inf[X_\alpha]$ in $P(\kappa)/I$, or is that just a very nice property of normal ideals?
Sep 14, 2012 at 13:52 history answered Sean Cox CC BY-SA 3.0