Subset of a P-ideal need not be a P-ideal

Let $$D_k=\{n\in\mathbb N; \text{prime factorization of $n$ has precisely $k$ prime factors}\}.$$ This is decomposition of $\mathbb N$ into infinitely many infinite pairwise disjoint subsets.$\newcommand{\I}{\mathcal I}\newcommand{\J}{\mathcal K}$

Let us define $\J=\{A\subseteq\mathbb N; A\text{ intersects only finitely many of sets }D_n\}$. Then $\J$ is an ideal and it is not a P-ideal. (To see this just take $A_n=D_n$.)

But every set in $\J$ has asymptotic density zero, so $\J\subseteq\I=\{A\subseteq\mathbb N; d(A)=0\}$. It is known that ideal consisting of sets having zero asymptotic density is a P-ideal. In fact, S. Solecki described large class of analytic P-ideals in a similar way, with some type of submeasure instead of asymptotic density; S.Solecki: Analytic Ideals. Bull. Symbolic Logic Volume 2, Number 3 (1996), 339-348; available here, projecteuclid, jstor.

The fact that each set $A\in\J$ has density zero is given as Corollary 2, in I. Niven: The asymptotic density of sequences; DOI:10.1090/S0002-9904-1951-09543-9, projecteuclid.

It's easy to construct such examples. First construct a sequence $A_\alpha$ ($\alpha < \omega_1$) of subsets of $\mathbb{N}$, satisfying $A_\alpha \subset^* A_\beta$ for all $\alpha < \beta < \omega_1$ (so $A_\alpha \setminus A_\beta$ is finite and $A_\beta \setminus A_\alpha$ is infinite). This is easy to do using a diagonalization argument.

At Martin's request, I'll outline the construction of such a sequence. Start with $A_0 = \emptyset$. At a successor step $\alpha+1$, let $A_{\alpha+1} = A_\alpha\cup B$, where $B$ is any infinite, co-infinite subset of $\mathbb{N}\setminus A_\alpha$. (For this we need to ensure that $A_\alpha$ is co-infinite for every $\alpha$; we'll see how to do this in the limit stage too.) Now for limit $\alpha$, suppose $\alpha_n$ is an increasing, cofinal sequence in $\alpha$, and let $A$ be the union $\bigcup A_{\alpha_n}$. For each $n$ choose a point $t_n\in A_{\alpha_n}$ such that $t_n$ is not in $A_{\alpha_m}$ for any $m < n$. Let $A_\alpha = A\setminus\{t_n | n\in\mathbb{N}\}$.

Now take $\mathcal{I}$ to be the ideal of subsets $A$ of $\mathbb{N}$ such that $A\subseteq^* A_\alpha$ for some $\alpha < \omega_1$, and, for some fixed limit ordinal $\beta < \omega_1$, let $\mathcal{K}_\beta$ be the ideal of subsets $A$ of $\mathbb{N}$ such that $A\subseteq^* A_\alpha$ for some $\alpha < \beta$. It's easy to see that $\mathcal{I}$ is a $P$-ideal. To see that $\mathcal{K}_\beta$ is not a $P$-ideal, consider a cofinal sequence $\beta_n$ in $\beta$; then the sequence $A_{\beta_n}$ has no $\subseteq^*$-upper bound in $\mathcal{K}_\beta$.

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.