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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

06C15 Complemented lattices, orthocomplemented lattices and posets
81P10 Logical foundations of quantum mechanics; quantum logic (quantum-theoretic aspects)
Full Text: EuDML