เหมาะสำหรับ $\mathbb{N}$ ด้วยทรัพย์สินบางอย่าง

Question

ปล่อย $\mathcal{I}$ เป็นคนในอุดมคติ $\mathbb{N}$ที่ประกอบด้วยเซต จำกัด ทั้งหมดและเซตอนันต์อย่างน้อยหนึ่งเซต กำหนดตัวกรอง

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

$\mathcal{F}$ มีตัวกรองโคฟินิทและดูเหมือนว่าถ้า $\mathcal{I}$ 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.

Answer 1

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.

Answer 2

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.