Abstract
A lattice L is said to be regularly embedded in a complete lattice K if there exists an isomorphic mapping of L into K which preserves all least upper and greatest lower bounds that exist in L. With the MacNeille completion by cuts every lattice can be regularly embedded in a complete lattice. And if the lattice is a Boolean algebra, then its completion by cuts is also a Boolean algebra. In general, however, this embedding does not preserve lattice structure; there are examples of distributive lattices whose completions by cuts are not even modular. On the other hand every lattice L can be embedded in the complete lattice of its ideals which satisfies every identity that holds in L. This embedding preserves greatest lower bounds but does not preserve infinite least upper bounds. Recently A. Horn has asked whether or not any distributive lattice can be regularly embedded in some complete distributive lattice. Here this question is answered in the negative. We show with an example that there exists a distributive lattice which cannot be regularly embedded in a complete modular lattice. What appears to be most important in the example is the lack of certain continuity laws. If we impose such a law on a lattice, then a structure preserving regular embedding is possible. Following [2] let us define a subset A of a lattice L to be a complete ideal if for every element aEL and every subset SCA for which US exists in L we have
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