The following little question has bugged me for a while.


Suppose $Z \subseteq \mathcal P(X)$. We say an ideal $I$ on $Z$ is *normal* when it is closed under diagonal unions, which means that if $\{ A\_x : x \in X \} \subseteq I$, then $\nabla\_x A\_x := \{ z \in Z : \exists x \in z ( z \in A\_x) \} \in I$. We say that $I$ is *fine* when for all $x \in X$, $\{ z \in Z : x \notin z \} \in I$. We say that $I$ is countably complete if it is closed under countable unions.


**Fact 1:** Normal + fine implies countably complete. See Proposition 1.5 [here](https://arxiv.org/pdf/1901.01160.pdf).


**Fact 2:** Fine does not imply countably complete. Let $Z$ be the collection of all finite subsets of an infinite set $X$, and let $I$ be the smallest fine ideal on $Z$.


**Fact 3:** Normal + countably complete does not imply fine. Let $Z$ be the set of all Dedekind cuts of the rationals, and let $I$ be a countably complete ideal on the reals like the Lebesgue null ideal. Then $I$ is normal by the regressive function characterization, since any regressive function on a non-null set takes some rational value on a non-null subset. But $I$ is not fine, since the set of cuts not containing a given rational is just the reals less than that rational, which is not null.


**Fact 4:** Fine + countably complete does not imply normal. It is not hard to show that the ideal of countable subsets of $\omega\_1$ is not normal.


**Question:** Can we strengthen Fact 1? Does normality imply countable completeness?



For ideals, does normal imply countably complete?
I think the answer is yes. It helped to take generic ultrapowers, somehow.


We may assume without loss of generality that $I$ is a normal ideal on $P(\lambda)$ where $\lambda$ is an infinite cardinal. Assume towards a contradiction that $I$ is countably incomplete. Then there is an $I$-positive set $S\subseteq P(\lambda)$ such that $S = \bigcup\_{n < \omega} A\_n$ where $A\_n$ is null for all $n < \omega$. Let $G$ be generic for the partial order of $I$-positive sets under inclusion and assume $S\in G$.


We work in $V[G]$ for a while. Let $j\_G : V\to M\_G$ be the generic ultrapower. Let $\tau = \{a\in M\_G : M\_G\vDash a\in [\text{id}]\_G\}$ be the extension of $[\text{id}]\_G$. Since $I$ is normal, $G$ is $V$-normal, which is equivalent to saying $\tau\subseteq j\_G[\lambda]$. Also $G$ is $V$-countably incomplete, which means that $M\_G$ is $\omega$-nonstandard. As a consequence, no structure in $M\_G$ can truly be an infinite wellorder. Since $\tau\subseteq j\_G[\lambda]$, $V[G]$ satisfies that $\tau$ is wellordered by the relation $\in^{M\_G}$. The canonical order of $[\text{id}]\_G$ in $M\_G$ codes this wellorder of $\tau$, and hence $\tau$ is finite (in $V[G]$). So there is a finite set $\sigma\subseteq \lambda$ such that $\tau = j\_G[\sigma]$. The finite set of ordinals $\sigma$ belongs to $V$. Therefore some $T\in G$ forces $[\text{id}]\_G = j\_G(\sigma)$.


Returning to $V$, the fact that $T$ forces $[\text{id}]\_G = j\_G(\sigma)$ means that for $I$-almost all $\upsilon\in T$, $\upsilon = \sigma$. In other words, $T$ is the union of $\{\sigma\}$ and an $I$-null set. It follows that $\{\sigma\}$ is $I$-positive. Since $\sigma\in T\subseteq S = \bigcup\_{n < \omega} A\_n$, there is some $n < \omega$ such that $\sigma\in A\_n$, and so $A\_n$ is $I$-positive, a contradiction.


