Editing Complete lattice (section)

=== Free complete lattices ===
The situation for complete lattices with complete homomorphisms is more intricate. In fact, free complete lattices generally do not exist. Of course, one can formulate a word problem similar to the one for the case of [[lattice (order)|lattices]], but the collection of all possible [[word problem (mathematics)|words]] (or "terms") in this case would be a [[proper class]], because arbitrary meets and joins comprise operations for argument sets of every [[cardinality]].

This property in itself is not a problem: as the case of free complete semilattices above shows, it can well be that the solution of the word problem leaves only a set of equivalence classes. In other words, it is possible that the proper classes of all terms have the same meaning and are thus identified in the free construction. However, the equivalence classes for the word problem of complete lattices are "too small," such that the free complete lattice would still be a proper class, which is not allowed.

Now, one might still hope that there are some useful cases where the set of generators is sufficiently small for a free, complete lattice to exist. Unfortunately, the size limit is very low, and we have the following theorem:

: The free complete lattice on three generators does not exist; it is a [[proper class]].

A proof of this statement is given by Johnstone.<ref>P. T. Johnstone, ''Stone Spaces'', Cambridge University Press, 1982; ''(see paragraph 4.7)''</ref> The original argument is attributed to [[Alfred W. Hales]];<ref>[[Alfred W. Hales|A. W. Hales]], ''On the non-existence of free complete Boolean algebras'', Fundamenta Mathematicae 54: pp.45-66.</ref> see also the article on [[free lattice]]s.