A complete and countable orthomodular lattice is atomic
Charles H. Randall · Proceedings of the American Mathematical Society · 1969
To prove the assertion in the title, let (L; <, ') be a countable orthomodular lattice. If L is not atomic, there exists a nonzero bC-L that does not majorize an atom. In L, let f<g<b. By the orthomodular identity g=fV(f'Ag) and necessarily (f'Ag) 00 Since (f'Ag) < g < b, (f'Ag) can not be an atom and there exists in L a nonzero d < (f'Ag). Again (f'Ag) = d V (d'A\f'Ag) and (d'/\f'Ag) 0O. Let h=fVd. Since g=fVdV(d'Af'Ag) is an orthogonal join of nonzero elements, it follows thatf<h<g< b.-Thus the interval [0, b] = {g| 0<g?b} is order dense-in-itself (i.e., if f<g in [0, b] there exists an h E [0, b ] such that f < h < g). Let C, by Zorn's Lemma, be a maximal chain in [0, b]. It follows that C is bounded, countable and order dense-in-itself. Hence C must be order isomorphic to the chain of all rational numbers between any pair of fixed distinct endpoints (see [1, p. 200]). It is well known that such a chain is not order complete. Therefore there exists a C'CC for which no VC' exists in C. If such an element VC' exists in L, it must be in [0, b], for b is an upper bound of C'. In that event, by the maximality of C, VC' must also be in C. Thus no VC' exists in L and L is not complete. Q.E.D.