# pseudodifference

## Pseudodifference of filter objects ★★

Author(s): Porton

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

I will call \emph{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. \href[See here for the formal definition of filter objects]{http://www.mathematics21.org/binaries/filters.pdf}. I will denote $(\operatorname{up} a)$ the filter corresponding to a filter object $a$. I will denote the set of filter objects (on $U$) as $\mathfrak{F}$.

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

\begin{problem} Which of the following expressions are pairwise equal for all $a, b \in \mathfrak{F}$ for each set $U$? (If some are not equal, provide counter-examples.) \begin{enumerate} \item $\bigcap^{\mathfrak{F}} \left\{ z \in \mathfrak{F} | a \subseteq b \cup^{\mathfrak{F}} z \right\}$;

\item $\bigcup^{\mathfrak{F}} \left\{ z \in \mathfrak{F} | z \subseteq a \wedge z \cap^{\mathfrak{F}} b = \emptyset \right\}$;

\item $\bigcup^{\mathfrak{F}} (\operatorname{atoms} a \setminus \operatorname{atoms} b)$;

\item $\bigcup^{\mathfrak{F}} \left\{ a \cap^{\mathfrak{F}} (U\setminus B) | B \in \operatorname{up} b \right\}$. \end{enumerate} \end{problem}

Keywords: filters; pseudodifference