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.

  • $\begingroup$ Thanks! I know that, if $L$ is a Boolean algebra, then (almost trivially) the set of complements of the elements of an ideal is a filter. Do you know what further condition a lattice must satisfy between being a pseudocomplemented lattice and being a Boolean algebra in order for this to work? Or is this true just for Boolean algebras? $\endgroup$ – Nagase Oct 28 '14 at 21:13
  • $\begingroup$ @Nagase: You’re welcome! I don’t know, I’m afraid, at least not off the top of my head; I’m a topologist, and what I know of lattice theory I’ve picked up in bits and pieces along the way. $\endgroup$ – Brian M. Scott Oct 28 '14 at 21:15
  • $\begingroup$ @BrianMScott - I was re-reading your answer and I have a doubt. Your conclusion states that every filter containing $U$ will contain a set that isn't a pseudo-complement; however, what is needed for a counter-example is the proposition that every filter containing $\neg U$ contains a set that isn't a pseudo-complement, right? $\endgroup$ – Nagase Oct 29 '14 at 0:30
  • $\begingroup$ @Nagase: The point is if $U$ is any open set at all (other then $\Bbb R$), any filter containing it will contain a set that isn’t a pseudo-complement. In particular, you can take $U$ to be $\neg V$ for some non-empty $V\in\tau$. $\endgroup$ – Brian M. Scott Oct 29 '14 at 0:34
  • $\begingroup$ @BrianMScott - Ahh, got it. That makes sense. Thanks a lot again! $\endgroup$ – Nagase Oct 29 '14 at 0:52

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