Let $L$ be a pseudocomplemented distributive lattice with $0$ and $1$, $I \subseteq L$ an ideal and set $F = \{\neg x \; | \; x \in I\}$, where $\neg x$ is the pseudocomplement of $x$. My question is: is $F$ a filter? It's clear that, for any two elements $x, y \in F$, their meet $x \wedge y$ is in $F$, but it's not entirely clear to me that, if $x \in F$ and $x \leq y$, then $y \in F$. In particular, this would seem to imply that every element that is greater than a pseudocomplement is the pseudocomplement of another element, yet I'm not entirely sure if this holds (in fact, my attempted at proving that $F$ is a filter stumbled precisely at this point).
I've been trying to think of counterexamples, but I'm unsure if I'm on the right track. I thought of using the structure $\langle \mathcal{P}(\mathbb{R}), \cap, \cup, \emptyset, \mathbb{R}, \neg \rangle$ as a counterexample, where $\neg X = \operatorname{int}(\mathbb{R} \setminus X)$. If this is a pseudocomplemented lattice, then, if we consider the order generated by inclusion, then taking the ideal generated by, say, the interval $(0, 2)$, and then taking the pseudocomplements of its elements won't generate a filter, as there will be many closed intervals which will be left outside of this set. Is this on the right track?
I would just take the lattice $\tau$ of open sets in $\Bbb R$ with the usual topology. Then for any $U\in\tau$ you do indeed have a pseudo-complement $\neg U=\operatorname{int}(\Bbb R\setminus U)=\Bbb R\setminus\operatorname{cl}U$. Note that all pseudo-compements are regular open sets.
If $U$ is a regular open set other than $\Bbb R$ itself, $\neg U$ is a non-empty regular open set. Let $(a,b)$ be any open interval contained in $\neg U$, fix $c\in(a,b)$, and let $V=U\cup(a,c)\cup(c,b)$. Then $U\subseteq V$, but $V$ is not regular open, since $c\in(\operatorname{int}\operatorname{cl}V)\setminus V$, so $V$ is not a pseudo-complement in $\tau$. Thus, every filter containing $U$ must contain a set that isn’t a pseudo-complement.