For ideals, does normal imply countably complete?

Question

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?

Answer 1

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.

Answer 2

Here's an attempt to show that normality implies countable completeness in the case where the underlying set $X$ is wellorderable and each member of $Z$ is infinite. We may suppose that $X$ is some limit ordinal $\gamma$. Suppose that $B_{i}$ $(i \in \omega)$ are subsets of $\mathcal{P}(\gamma)$, and let $I$ be the normal ideal generated by $\{ B_{i} : i < \omega\}$. We want to show that $\bigcup_{i \in \omega}B_{i}$ is in $I$.

The idea for what follows is that we want to build a tree of diagonal unions corresponding to the tree of descending sequences from $\gamma$ (except that each sequence also says how much longer it will continue), with terminal sequences of length $n$ associated to $B_{n}$.

Define sets $A_{\alpha, n,m} \in I$ for $(\alpha, n, m) \in (\gamma + 1) \times \omega \times \omega$ recursively in $m$ and simultaneously in $\alpha$ and $n$. Each $A_{\alpha,n,0}$ is $B_{n}$. When $m > 1$ and $\alpha \geq m$, $A_{\alpha,n,m}$ is the diagonal union with sets $A_{\beta,n+1,m-1}$ in positions $\beta < \alpha$, and the emptyset in positions indexed in $\gamma \setminus \alpha$. When $\alpha < m$ we let $A_{\alpha, n,m}$ be the emptyset.

Note that if $\alpha < \gamma$, then $A_{\alpha, n,m} \subseteq A_{\gamma, n,m}$.

Finally, let $f \colon \gamma \to \omega$ take all values cofinally often, and let $C = \bigtriangledown_{\alpha < \gamma} A_{\alpha, 0, f(\alpha)}$. We want to see that $C = \bigcup_{i \in \omega}B_{i}$.

Fix $z$ in some $B_{n}$. Let $\langle \alpha_{i} : i < n\rangle$ be a descending sequence of elements of $z$, and let $\beta \in [\alpha_{0}, \gamma)$ be such that $f(\beta) = n$. Then $z$ is in each of the sets $A_{\alpha_{n},n,0}$, $A_{\alpha_{n-1},n-1,1}$ and so on up to $A_{\alpha_{0},0,n}$. This implies that $z \in A_{\beta, 0,n}$, so $z \in C$.

Answer 3

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). $\square$