Principal filters are center – solved

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.

[1] Victor Porton. Funcoids and Reloids.

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