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Projection and covering in a set with orthogonality. (English) Zbl 0629.06008
Let $$\Omega$$ be a set with orthogonality $$\perp$$ and $$S=\{A^{\perp}|$$ $$A\subseteq \Omega \}$$. Assume there exists $$0\in \Omega$$ with $$0\perp x$$ for all $$x\in \Omega$$ and $$\{x\}^{\perp \perp}$$ is an atom in the lattice S for all $$x\in \Omega -\{0\}$$. The author proves the equivalence of the following statements where A, $$A_ 1$$, $$A_ 2$$ are elements of S and $$x\in \Omega:$$ (i) if $$x\not\in A$$ then $$A\vee \{x\}^{\perp \perp}$$ covers A, (ii) if $$x\not\in A\cup A^{\perp}$$ then $$x\in A_ 1\vee A_ 2$$ for some atoms $$A_ 1\subset A$$, $$A_ 2\subset A^{\perp}$$, (iii) if $$A_ 1\perp A_ 2$$, $$A_ 1\neq \{0\}$$, $$A_ 2\neq \{0\}$$, $$x\in A_ 1\vee A_ 2$$, $$x\not\in A_ 1$$, $$x\not\in A_ 2$$ then there exist $$x_ i\in A_ i$$ with $$x\in \{x_ 1\}^{\perp \perp}\vee \{x_ 2\}^{\perp \perp}$$, (iv) if $$x\not\in A\cup A^{\perp}$$ then $$A\cap (A^{\perp}\vee \{x\}^{\perp \perp})$$, $$A^{\perp}\cap (A\vee \{x\}^{\perp \perp})$$ are atoms in S.
Reviewer: G.Kalmbach

##### MSC:
 06C15 Complemented lattices, orthocomplemented lattices and posets 81P10 Logical foundations of quantum mechanics; quantum logic (quantum-theoretic aspects)
##### Keywords:
projection; covering; orthogonality; atom; covers
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