Enough homomorphisms to separate its elements

Question

For a lattice $L$ what does the statement mean that there are enough homomorphisms $L\to \{0,1\}$ to separate its elements? What exactly is meant by "separating"?

Answer 1

Given a lattice $L$, let $H$ be the set of lattice homomorphisms $L \to \{ 0, 1 \}$. We have a map $L \to \{ 0, 1 \}^H$ given by $x \mapsto (f \mapsto f (x))$, and there are enough lattice homomorphisms $L \to \{ 0, 1 \}$ to separate the elements of $L$ precisely if $L \to \{ 0, 1 \}^H$ is injective. In more elementary terms, the condition says:

• Given $x$ and $y$ in $L$, if $f (x) = f (y)$ for every lattice homomorphism $f : L \to \{ 0, 1 \}$, then $x = y$.

Since $\{ 0, 1 \}$ is a distributive lattice, if there are enough homomorphisms $L \to \{ 0, 1 \}$ to separate elements of $L$, then $L$ must be a distributive lattice. In fact, the converse is also true: if $L$ is a distributive lattice, then there are enough homomorphisms $L \to \{ 0, 1 \}$ to separate elements of $L$. This is fundamental to Stone duality for distributive lattices.