by lattice, i mean definition 2 at http://en2.wikipedia.org/wiki/Complete+lattice [Broken] .

this is probably ill-posed but here goes nothing...

is there a nonempty lattice such that the maximal element of the lattice is the whole lattice?

so what i mean by a maximal element is an element b such that for all a in the lattice, a <= b. is there a lattice such that such a b is the whole lattice?

i'm sure this has been tried before but perhaps one can approach set theory through lattices as defined in definition 2 of http://en2.wikipedia.org/wiki/Complete+lattice [Broken].

two axioms of lattice set theory would be
1. there is an element in the lattice, Ø, such that for all x,
x^Ø=Ø and x v Ø=x and
2. there is an element in the lattice, U, such that for all x,
x^U=x and x v U=U.

another would be that x^y=y^x and x v y=y v x.

i'm suspecting there might be a problem when one allows x to be U or Ø in the two axioms:
1. (x=U). U^Ø=Ø and U v Ø = U.
2. (x=Ø). Ø^U=Ø and Ø v U = U. ok, i guess there's no contradiction so far.

i'm trying to avoid fuzzy logic, if possible, at least for right now.

one of the main issues is how to restate a version of the subsets axiom. i think the definiion of subset would have to be that x is a subset of y if and only if x<=y which means x v y=y. i'd like to have a subsests axiom so that given a y and well-formed-formula (wff) p, there is an x such that z ∈ x iff (z ∈ y and p(z)). just a thought: in two-valued logic, p(z) is either true or false. maybe i can switch and to meet, ^, and define p(z) to be U if p(z) is true and Ø if p(z) is false. the other problem will be to define ∈ . i'd want it to be defined in terms of meet and join and so that x is a subset of y if and only if (z ∈ x implies z ∈ y). one random candidate is that x ∈ y would be the same as x<y which means that (x!=y and x<=y). well, whatever ∈ means, S(y,p):={z ∈ y : p(z)} could be defined so that z ∈ S(y,p) iff (z ∈ y)^p(z) or something...

however i handle the subsets axiom, i want to avoid russell's paradox, of course. that would be the case of s:=S(U,p) where p(z) says z ! ∈ z when one asks if s ∈ s.