October 31, 2009 / porton

Principal filters are center – solved

Theorem 1 If ${\mathfrak{F}}$ is the set of filter objects on a set ${U}$ then ${U}$ is the center of the lattice ${\mathfrak{F}}$ . (Or equivalently: The set of principal filters on a set ${U}$ is the center of the lattice of all filters on ${U}$ .)

Proof: I will denote ${Z (\mathfrak{F})}$ the center of the lattice ${\mathfrak{F}}$ . I will denote ${\mathrm{atoms}^{\mathfrak{A}} a}$ the set of atoms of a lattice ${\mathfrak{A}}$ under its element ${a}$ .

Let ${\mathcal{X} \in Z (\mathfrak{F})}$ . Then exists ${\mathcal{Y} \in Z (\mathfrak{F})}$ such that ${\mathcal{X} \cap^{\mathfrak{F}} \mathcal{Y} = \emptyset}$ and ${\mathcal{X} \cup^{\mathfrak{F}} \mathcal{Y} = U}$ . Consequently, there are ${X \in \mathrm{up} \mathcal{X}}$ such that ${X \cap^{\mathfrak{F}} \mathcal{Y} = \emptyset}$ ; we have also ${X \cup^{\mathfrak{F}} \mathcal{Y} = U}$ . Suppose ${X \supset \mathcal{X}}$ . Then (because for ${\mathfrak{F}}$ is true the disjunct propery of Wallman, see [1]) exists ${a \in \mathrm{atoms}^{\mathfrak{F}} X}$ such that ${a \notin \mathrm{atoms}^{\mathfrak{F}} \mathcal{X}}$ . We can conclude also ${a \notin \mathrm{atoms}^{\mathfrak{F}} \mathcal{Y}}$ . Thus ${a \notin \mathrm{atoms}^{\mathfrak{F}} ( \mathcal{X} \cup^{\mathfrak{F}} \mathcal{Y})}$ and consequently ${\mathcal{X} \cup^{\mathfrak{F}} \mathcal{Y} \neq U}$ what is a contradiction. We have ${\mathcal{X} = X \in \mathscr{P} U}$ .

Let now ${X \in \mathscr{P} U}$ . Then ${X \cap (U \setminus X) = 0}$ and ${X \cup (U \setminus X) = U}$ . Thus ${X \cap^{\mathfrak{F}} (U \setminus X) = \bigcap^{\mathfrak{F}} \left\{ X \cap (U \setminus X) \right\} = \emptyset}$ ; ${X \cup^{\mathfrak{F}} (U \setminus X) = \bigcap^{\mathfrak{F}} (\mathrm{up} X \cap \mathrm{up} (U \setminus X)) = \bigcap^{\mathfrak{F}} \left\{ U \right\} = U}$ (used formulas from [1]). We have shown that ${X \in Z (\mathfrak{F})}$ . $\Box$