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.)

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

3 thoughts on “Open problem: Pseudodifference of filters

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