# Ideal on $\mathbb{N}$ with certain property

Let $$\mathcal{I}$$ be an ideal on $$\mathbb{N}$$ that contains all finite sets and at least one infinite set. Define a filter

$$\mathcal{F}:=\{D\subseteq\mathbb{N}\mid \forall A\in\mathcal{I},A\cap D^{c}\text{ is finite, or equivalently} A\subseteq^{*}D\}$$.

$$\mathcal{F}$$ contains the cofinite filter, and it seems that if $$\mathcal{I}$$ is prime then $$\mathcal{F}$$ does not contain anything else. Does the converse hold? In other words, let us say an ideal has property P if the corresponding filter is the cofinite filter. Is P the same as being prime? Or is there simple characterization of P?

Someone suggested that this is same as asking for $$\mathcal{E}\subseteq(\mathcal{P}_{coinf}(\mathbb{N}),\subseteq^{*})$$ which is unbounded under $$\subseteq^{*}$$ and generates a proper non-prime ideal. I found that I know nothing about this poset. What is its cofinal type? What is its relation with other posets such as $$(\mathbb{N}^{\mathbb{N}},<^{*})$$?

Background: I was thinking if we define a topology on $$\mathbb{N}\cup\{\infty\}$$ by requiring certain sequences converge to $$\infty$$, will there be more (and which) sequences converging to $$\infty$$ than we expected. Also see this question.

• Ideal with your property P are called tall ideals. They've been extensively studied. An early source is Adrian Mathias's paper "Happy families"; a fairly recent survey of this and lots of other properties of ideals is Michael Hrusak's "Combinatorics of filters and ideals". Aug 20, 2020 at 2:22
• @AndreasBlass Thanks so much. Actually I noticed the second paper but didn't even scan through it carefully...Could you make this an answer so that I can accept it? Aug 20, 2020 at 2:41

Given $$P_1,P_2$$ non-principal prime ideals on $$\mathbb{N}$$ with $$P_1\neq P_2$$, let $$\mathcal{I}= P_1\cap P_2$$. Then $$\mathcal{I}$$ is an ideal containing all finite sets, but not prime (as there must be some $$A\subseteq \mathbb{N}$$ with $$A\notin P_1, A^c\notin P_2$$).
However $$\mathcal{I}$$ satisfies property P: Given any $$D$$ not cofinite, we may partition $$D^c$$ into $$4$$ infinite pieces: $$D_{11,}D_{12},D_{21},D_{22}$$. Then $$D_{i1}\cup D_{i2}\in P_1$$ for some $$i$$ and $$D_{1j}\cup D_{2j}\in P_2$$ for some $$j$$. Thus $$D_{ij}\in \mathcal{I}$$ and $$D_{ij}\cap D^c=D_{ij}$$ is infinite.
• $\mathcal{I}$ is required to contain all finite sets. Aug 20, 2020 at 0:07
• @1830rbc03 Thanks, good point - I have fixed this by replacing elements of $\mathbb{N}$ with non-principal ultrafilters on $\mathbb{N}$.The same argument works, but now $\mathcal{I}$ contains all finite sets.
Let $$X$$ be a maximal almost disjoint family of subsets of $$\mathbb{N}$$, and let $$\mathcal{I}$$ be the ideal generated by $$X$$. Then $$\mathcal{F}$$ will be the cofinite filter: if $$D\in\mathcal{F}$$ then $$D^c$$ is almost disjoint from every element of $$X$$, and thus must be finite by maximality of $$X$$. However, $$\mathcal{I}$$ is not prime. For instance, if you take two disjoint countably infinite subfamilies $$Y,Z\subset X$$, then by a simple diagonalization argument you can construct $$A\subset\mathbb{N}$$ which almost contains every element of $$Y$$ and is almost disjoint from every element of $$Z$$. Then $$A\not\in\mathcal{I}$$ since every element of $$\mathcal{I}$$ has infinite intersection with only finitely many elements of $$X$$, and $$A^c\not\in\mathcal{I}$$ for the same reason.