# Ideals on $\mathbb N$ and large sets that have small intersection

Let $$\mathcal I$$ be a (non-principal) ideal of subsets of $$\mathbb N$$. Suppose that every family $$\mathcal{A} \subset \wp(\mathbb N)\setminus \mathcal I$$ with the following property is countable: $$A,B\in \mathcal A, A\neq B \Rightarrow A\cap B \in \mathcal I.$$

Let us observe that maximal ideals have this property as only the empty family or the singletons meet this requirement.

Are there further examples of ideals with this property?

(As observed by Leonetti every such ideal must be non-meagre when regarded as a subset of the Cantor set.)

• "only the empty family": you mean "only the empty family or singletons". By the way, what you call "family" is usually called a "subset". Of course, every set can be thought of as a family $(a)_{a\in A}$... – YCor Sep 22 '18 at 16:50
• You have all ideals $I$ such that $P(N)/I$ is countable (but this forces $P(N)/I$ finite). – YCor Sep 22 '18 at 16:51
• By the way this is purely ring-theoretic: in a ring $R$ (say associative unital commutative), you considers ideals $I$ and subsets $\mathcal{A}\subset R\smallsetminus I$ such that for all $a\neq b\in\mathcal{A}$ we have $ab\in I$, and require that all such subsets are countable. In a Boolean algebra $a,b\notin I,a-b\in I$ implies $ab\notin I$, so this forces $\mathcal{A}$ to embed into $A/I$. Beyond the Boolean case, this applies when $A/I$ is reduced. – YCor Sep 22 '18 at 16:54

The denumerable atomless Boolean algebra $$A$$ can be isomorphically embedded in $$\wp(\omega)/\mathsf{fin}$$. The identity on $$A$$ can be extended to a homomorphism from $$\wp(\omega)/\mathsf{fin}$$ into the completion of $$A$$. This gives a homomorphic image of $$\wp(\omega)/\mathsf{fin}$$, hence also of $$\wp(\omega)$$ itself, in which every disjoint subset is countable. So the kernel of the homomorphism is another ideal of the type indicated.

I'm not sure the following example is different from the ones already given, but the description is different, and so I hope some readers might find it useful. Fix a countable family $$\mathcal F$$ of non-principal ultrafilters on $$\mathbb N$$, and let $$I$$ be the ideal of sets that are in none of the ultrafilters in $$\mathcal F$$. Then $$I$$ is not maximal, because if $$X\subseteq\mathbb N$$ is in some but not all of the ultrafilters in $$\mathcal F$$ then neither $$X$$ nor its complement is in $$I$$. (Almost as easily, $$I$$ is not the intersection of finitely many maximal ideals.)

To prove that $$I$$ has the countability property in the question, consider any uncountable family $$\mathcal A$$ of sets not in $$I$$, Then each set in $$\mathcal A$$ is in at least one of the ultrafilters in $$\mathcal F$$. But $$\mathcal F$$ is countable and $$\mathcal A$$ is not, so there must be two (in fact uncountably many) sets $$A,B\in\mathcal A$$ belonging to the same ultrafilter $$U\in\mathcal F$$. Then $$A\cap B$$ is in $$U$$ and therefore not in $$I$$.

In topological language: for any closed subset, $$F$$, of $$\beta\mathbb{N}\setminus\mathbb{N}$$ the family $$I_F=\{A:A^*\cap F=\emptyset\}$$ is an ideal. Your property translates into: the closed set $$F$$ is a ccc space.

The example by Monk can be made in this way too: take a map $$s$$ from $$\beta\mathbb{N}\setminus\mathbb{N}$$ onto the Cantor set $$C$$ and take $$F$$ such that $$s:F\to C$$ is irreducible and onto, then $$F$$ is sparable and hence ccc.

The example by Blass corresponds to the closure of a countable set in $$\beta\mathbb{N}\setminus\mathbb{N}$$, again a separable subspace.

In general: take any compact ccc space, $$X$$, of weight at most $$\mathfrak{c}$$ and embed it into the Tychonoff cube $$[0,1]^\mathfrak{c}$$. There is a continuous map $$\sigma$$ from $$\beta\mathbb{N}$$ onto that cube. Now take $$F$$ such that $$\sigma:F\to X$$ is irreducible and onto; then $$F$$ is ccc. For a non-separable example let $$X$$ be the Stone space of the measure algebra.

I believe you can use forcing to cook up many examples: Let $$P=Add(\omega, 1)$$. This is the forcing that adds a subset of $$\mathbb{N}$$. Let $$\dot {U}$$ be a $$P$$-name of a non-principal ultrafilter. We can choose such a name with the property that for some infinite co-infinite $$X\subset \omega$$, $$\not\Vdash_P \check{X}\in \dot{U}$$ nor $$\not\Vdash_P \check{X^c}\in \dot{U}$$.

Define $$I=\{X\subset \omega: \Vdash_P X\not\in \dot{U}\}$$. This ideal is clearly not principal and is not maximal either (neither $$X$$ nor $$X^c$$ is in). Given any $$\{A_i\in P(\omega)-I: i<\omega_1\}$$, we find $$p_i\in P$$ such that $$p_i\Vdash_P A_i\in \dot{U}$$. But $$P$$ is countable so there exists $$i\neq j$$ such that $$p_i=p_j\Vdash A_i, A_j\in \dot{U}$$ hence $$p_i=p_j\Vdash_P A_i\cap A_j\in \dot{U}$$, which means $$A_i\cap A_j\not\in I$$. Any forcing with countable chain condition works here.

Another class of examples comes from the ideal of $$m$$-measure zero sets where $$m$$ is a finitely additive atomless probability measure on the power set of integers. Such measures can be constructed using Hahn-Banach theorem, ultralimits etc. For example, if $$U$$ is a non principal ultrafilter on the set of integers, then the $$U$$-density function $$m(A) = \lim_{n \in U} |A \cap [0, n]|/n$$ defines such a measure.