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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 $(A_n)_{ninmathbb N}$ of sets from $I$, there is an $AinI$ such that $A_nsubset^*A$ for all $n$'s. (The notation $Asubset^*B$ means that $Asetminus B$ 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 $JsubseteqI$, where $I$ is a P-ideal but $J$ is not a P-ideal.

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 $A_{alpha+1} = A_alphacup B$, 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 $t_nin A_{alpha_n}$ such that $t_n$ is not in $A_{alpha_m}$ for any $m < n$. Let $A_alpha = Asetminus{t_n | ninmathbb{N}}$.

Now take $mathcal{I}$ to be the ideal of subsets $A$ of $mathbb{N}$ such that $Asubseteq^* A_alpha$ 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 $Asubseteq^* A_alpha$ 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 $$D_k={ninmathbb N; text{prime factorization of $n$ has precisely $k$ prime factors}}.$$
This is decomposition of $mathbb N$ into infinitely many infinite pairwise disjoint subsets.

Let us define $J={Asubseteqmathbb N; Atext{ intersects only finitely many of sets }D_n}$. 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 $JsubseteqI={Asubseteqmathbb N; d(A)=0}$. 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 $AinJ$ 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.

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.