On semicontinuous linear lattices

Abstract

Applying spectral theory, we proved that a linear lattice is continuous if and only if it is semicontinuous and uniformly complete. In this paper we give another proof without use of spectral theory. Let L be a linear lattice. A sequence of elements xv E L (v= 1, 2, ) is called a uniform Cauchy sequence if there is a E L such that for any e>O we can find v0 for which IxM-xVI v0. L is said to be uniformly complete if every uniform Cauchy sequence is convergent. We can easily prove that every uniformly complete linear lattice is Archimedean, as done in [1]. A linear lattice L is said to be semicontinuous if every element x E L is normalable, i.e. if {x}' is a normal manifold for any x E L. In [1] we proved the THEOREM. A linear lattice is continuous if and only if it is semicontinuous and uniformly complete. We used spectral theory to prove it. In this paper we will give another proof without use of spectral theory. Let L be a linear lattice. If L is continuous, then L is semicontinuous by Theorem 6.15 of [2] and uniformly complete by Theorem 3.3 of [2]. Thus we will prove the converse. We suppose that L is semicontinuous and uniformly complete. First we prove that if a sequence 0?x4TL1 is bounded, then there is z E L such that [xv]xt1 1[z]x for x>O. If x_<k for v= 1 25 ... then setting = I ([Xv][xv-1])k for n = 1, 2,5 * * v=1 where x0=0, we obtain a uniform Cauchy sequence Yn (n= 1, 2, 5 Since L is uniformly complete by assumption, there exists z E L such that YnTn41z. Then [yn]xTn1=l[z]x for x_0 by Theorem 5.26 of [2] because Received by the editors October 14, 1971. AMS 1970 subject classifications. Primary 46A40.