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I have proved this conjecture:

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 ${X \in Z (\mathfrak{F})}$. $\Box$

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

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