TITLE: Subset of a P-ideal need not be a P-ideal
QUESTION [6 upvotes]: I was looking for examples showing that subset of a P-ideal is not necessary. I will post below a counterexample I was able to find. (I hope it is correct.) But I'd be glad to see other simple (or interesting) examples.$\newcommand{\I}{\mathcal I}\newcommand{\J}{\mathcal K}$
An ideal $\I$ of subsets of some set $X$ is said to be a P-ideal if,
for every sequence $(A_n)_{n\in\mathbb N}$ of sets from $\I$, there is an $A\in\I$ such that $A_n\subset^*A$ for all $n$'s. (The notation $A\subset^*B$ means that $A\setminus B$ is finite.)
Several equivalent reformulations of this property can be obtained relatively easily.
Some authors work with dual notion called P-filter. A filter $\mathcal F$ is P-filter if and only if every countable system of sets from $\mathcal F$ has a pseudointersection in $\mathcal F$. Ultrafilters with this property are called p-points.
I am mostly interested in P-ideals on countable sets, so we can w.l.o.g assume $X=\mathbb N$.

Question: What are some examples of ideals $\I$, $\J$ such that $\J\subseteq\I$, where $\I$ is a P-ideal but $\J$ is not a P-ideal.

REPLY [2 votes]: Let $\mathcal{I}$ be the ideal of asymptotic density zero sets and $\mathcal{K}$ be the ideal of Banach density zero sets. Then it is well known that $\mathcal{K} \subseteq \mathcal{I}$, $\mathcal{I}$ is a P-ideal, and $\mathcal{K}$ is not a P-ideal.