Question
A continuation/followup of my other question: Is the class of all equivalence relations (over an arbitrary domain), ordered by logical implication (or inclusion, if you prefer), a bounded lattice?
Motivation
Clearly, there are many, many subtle gradations of "sameness," but the intuition behind two things being "the same" is fundamentally that of an equivalence relation. The question becomes whether or not it is possible to specify the degree to which two objects are the same.
With this in mind, it seems reasonable to describe an ordering of equivalence relations which includes every shade of "same" and specifies the relationship between them. Naturally, this must be a partial ordering since there are equivalence relations which are logically independent of one another.
Reasoning
The reason I am drawn towards "lattice," and in particular "bounded" lattice, is that for any two equivalence relations $R_1$ and $R_2$, we can specify a third relation $R_3$ such that $\forall x,y.R_3(x,y)\iff(R_1(x,y)\land R_2(x,y))$, which is reminiscent of the "greatest lower bound" property of lattices. Furthermore, there is a least relation, the empty relation, which vacuously implies all others, and a greatest relation, "everything is the same," which is implied by all others.
My doubts stem from the sheer enormity of the scope of this claim (I am talking about all equivalence relations over any universe), and the fact that I am very tired and haven't had as much time to think it through as I would like.
For any set $X$, the equivalence relations on $X$ do form a bounded lattice (in fact, a complete one). Indeed, the top element is $X \times X$, the bottom element is $\Delta_X := \{(x,x)|x \in X\}$, the meet is the intersection, and the join is the transitive closure of the union (the union is already reflexive and symmetric, but not necessarily transitive).