Let $latex {U}&fg=000000$ is a set. A filter $latex {\mathcal{F}}&fg=000000$ (on $latex {U}&fg=000000$) is a non-empty set of subsets of $latex {U}&fg=000000$ such that $latex {A, B \in \mathcal{F} \Leftrightarrow A \cap B \in \mathcal{F}}&fg=000000$. Note that unlike some other authors I do not require $latex {\emptyset \notin \mathcal{F}}&fg=000000$.

I will call *the set of filter objects* the set of filters ordered reverse to set theoretic inclusion of filters, with principal filters equated to the corresponding sets. See here for the formal definition of filter objects. I will denote $latex {(\mathrm{up} a)}&fg=000000$ the filter corresponding to a filter object $latex {a}&fg=000000$. I will denote the set of filter objects (on $latex {U}&fg=000000$) as $latex {\mathfrak{F}}&fg=000000$.

I will denote $latex {(\mathrm{atoms} a)}&fg=000000$ the set of atomic lattice elements under a given lattice element $latex {a}&fg=000000$. If $latex {a}&fg=000000$ is a filter object, then $latex {(\mathrm{atoms} a)}&fg=000000$ is essentially the set of ultrafilters over $latex {a}&fg=000000$.

**Problem** Which of the following expressions are pairwise equal for all $latex {a, b \in \mathfrak{F}}&fg=000000$ for each set $latex {U}&fg=000000$? (If some are not equal, provide counter-examples.)

- $latex {\bigcap^{\mathfrak{F}} \left\{ z \in \mathfrak{F} | a \subseteq b \cup^{\mathfrak{F}} z \right\}}&fg=000000$;
- $latex {\bigcup^{\mathfrak{F}} \left\{ z \in \mathfrak{F} | z \subseteq a \wedge z \cap^{\mathfrak{F}} b = \emptyset \right\}}&fg=000000$;
- $latex {\bigcup^{\mathfrak{F}} (\mathrm{atoms} a \setminus \mathrm{atoms} b)}&fg=000000$;
- $latex {\bigcup^{\mathfrak{F}} \left\{ a \cap^{\mathfrak{F}} (U\setminus B) | B \in \mathrm{up} b \right\}}&fg=000000$.

## 3 thoughts on “Open problem: Pseudodifference of filters”