Weak ideal and its dual

Question

Let $(P, \leq)$ be a partially ordered set with an order reversing involution $^c$ i.e. $x\leq y\Rightarrow y^c\leq x^c$.

A subset $I$ of $P$ is a weak ideal, if the following conditions hold:

1. for any $x\in P$, either $x\in I$ or $x^c\in I$.
2. for every $x \in I$, any $y\in P$ and $y \le x$ implies that $y\in I$.

A filter $J$ of $P$ is a set of $P$ with following properties:

1. for any $x\in P$, either $x\in J$ or $x^c\in J$.
2. for every $y \in J$, any $x\in P$ and $y \le x$ implies that $x\in J$.

In particular I am interested in the case $P$ is the power set of a finite set. It seems that in this case every weak ideal is filter and vice versa.

Or we need some conditions to add to $P$ in order to have an equivalent definitions.

Answer 1

If $P$ is the power set of a finite set $X$ of size $n$ we can define $I$ to be all subsets of size $\lfloor \frac{n}{2}\rfloor $ or less. This is a weak ideal (if $^c$ is complementation) as we cannot have two disjoint sets $x$ and $x^c$ both of size $> \lfloor \frac{n}{2}\rfloor $, while their union is size $n$.

And a set that is a subset of a "small" set is also small.

But his $I$ is not a (weak) filter: Any filter contains $X$ and that is not in $I$. So ideals need not be filters, nor the other way around.

What does hold is that the notions are dual:

$I$ is a weak ideal iff $\{x^c : x\in I\}$ is a (weak) filter on $P$.

This follows from $x \to x^c$ being an order reversing involution. So we can always reduce theorems on filters to dual theorems on ideals and vice versa.