TITLE: For ideals, does normal imply countably complete?
QUESTION [14 upvotes]: 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.
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?

REPLY [2 votes]: Throughout, let $\kappa$ be a cardinal and let $X=\kappa$.
Here is a way to cover the case left open by Paul Larson.  Let $\{S_{n}\}_{n<\omega}\subseteq I$ be an arbitrary countable collection from $I$.  Let $T_{n}=\{z\in S_{n}\, :\, |z|<\aleph_0\}$.  Since $T_{n}\subseteq S_{n}\in I$ and $I$ is an ideal, we have $T_{n}\in I$.  If we knew that $T_{\infty}=\bigcup_{n<\omega}T_n\in I$, then by taking the union of $T_{\infty}$ with the set in Paul's answer we would be done.
We will prove the following (stronger) claim about possibly uncountable unions.
Claim:  Let $\{T_{\alpha}\}_{\alpha<\kappa}\subseteq I$.  If $z\in T_{\alpha}$ implies $z\in \mathscr{P}^{<\omega}(X)$ (i.e., $z$ is finite) for each $\alpha$, then $\bigcup_{\alpha<\kappa}T_{\alpha}\in I$.
Proof.  The claim is trivial if $\kappa$ is finite, so hereafter we assume $\kappa$ is infinite.  Working by induction on $\kappa$, suppose the claim is true for all (normal ideals on power sets of) smaller cardinals.
For each ordinal $\beta<\kappa$, let $T_{\alpha}(\beta)=\{z\in T_{\alpha}\, :\, z\in \mathscr{P}^{<\omega}(\beta)\}$. Since $\lambda=|\mathscr{P}^{<\omega}(\beta)|<\kappa$, we can fix an map $f\colon \lambda\to \kappa$ such that if $z\in \bigcup_{\alpha<\kappa}T_{\alpha}(\beta)$ then $z\in T_{f(\gamma)}(\beta)$ for some $\gamma<\lambda$.
Now, by our inductive hypothesis, the normal ideal generated by $\{T_{f(\gamma)}(\beta)\}_{\gamma<\lambda}$ contains
$$T_{\infty}(\beta)=\bigcup_{\gamma<\lambda}T_{f(\gamma)}(\beta)=\bigcup_{\alpha<\kappa}T_{\alpha}(\beta).$$
Hence, $T_{\infty}(\beta)\in I$ too.  The diagonal union of these sets (as indexed by $\beta$, possibly shifted by $1$) is $\bigcup_{\alpha<\kappa}T_{\alpha}$ (possibly after throwing in $\{\emptyset\}$, as needed). $\square$