### Principal filters are center – solved

I have proved this conjecture:

Theorem 1 If $latex {\mathfrak{F}}&fg=000000$ is the set of filter objects on a set $latex {U}&fg=000000$ then $latex {U}&fg=000000$ is the center of the lattice $latex {\mathfrak{F}}&fg=000000$. (Or equivalently: The set of principal filters on a set $latex {U}&fg=000000$ is the center of the lattice of all filters on $latex {U}&fg=000000$.)

Proof: I will denote $latex {Z (\mathfrak{F})}&fg=000000$ the center of the lattice $latex {\mathfrak{F}}&fg=000000$. I will denote $latex {\mathrm{atoms}^{\mathfrak{A}} a}&fg=000000$ the set of atoms of a lattice $latex {\mathfrak{A}}&fg=000000$ under its element $latex {a}&fg=000000$.

Let $latex {\mathcal{X} \in Z (\mathfrak{F})}&fg=000000$. Then exists $latex {\mathcal{Y} \in Z (\mathfrak{F})}&fg=000000$ such that $latex {\mathcal{X} \cap^{\mathfrak{F}} \mathcal{Y} = \emptyset}&fg=000000$ and $latex {\mathcal{X} \cup^{\mathfrak{F}} \mathcal{Y} = U}&fg=000000$. Consequently, there are $latex {X \in \mathrm{up} \mathcal{X}}&fg=000000$ such that $latex {X \cap^{\mathfrak{F}} \mathcal{Y} = \emptyset}&fg=000000$; we have also $latex {X \cup^{\mathfrak{F}} \mathcal{Y} = U}&fg=000000$. Suppose $latex {X \supset \mathcal{X}}&fg=000000$. Then (because for $latex {\mathfrak{F}}&fg=000000$ is true the disjunct propery of Wallman, see [1]) exists $latex {a \in \mathrm{atoms}^{\mathfrak{F}} X}&fg=000000$ such that $latex {a \notin \mathrm{atoms}^{\mathfrak{F}} \mathcal{X}}&fg=000000$. We can conclude also $latex {a \notin \mathrm{atoms}^{\mathfrak{F}} \mathcal{Y}}&fg=000000$. Thus $latex {a \notin \mathrm{atoms}^{\mathfrak{F}} ( \mathcal{X} \cup^{\mathfrak{F}} \mathcal{Y})}&fg=000000$ and consequently $latex {\mathcal{X} \cup^{\mathfrak{F}} \mathcal{Y} \neq U}&fg=000000$ what is a contradiction. We have $latex {\mathcal{X} = X \in \mathscr{P} U}&fg=000000$.

Let now $latex {X \in \mathscr{P} U}&fg=000000$. Then $latex {X \cap (U \setminus X) = 0}&fg=000000$ and $latex {X \cup (U \setminus X) = U}&fg=000000$. Thus $latex {X \cap^{\mathfrak{F}} (U \setminus X) = \bigcap^{\mathfrak{F}} \left\{ X \cap (U \setminus X) \right\} = \emptyset}&fg=000000$; $latex {X \cup^{\mathfrak{F}} (U \setminus X) = \bigcap^{\mathfrak{F}} (\mathrm{up} X \cap \mathrm{up} (U \setminus X)) = \bigcap^{\mathfrak{F}} \left\{ U \right\} = U}&fg=000000$ (used formulas from [1]). We have shown that $latex {X \in Z (\mathfrak{F})}&fg=000000$. $latex \Box&fg=000000$

This theorem may be generalized for a wider class of filters on lattices than only filters on lattices of a subsets of some set.