Subalgebra lattice

Remarks

The intersection of any family of subuniverses is a subuniverse, so $\operatorname{Sub} \mathbf{A}$ is a complete lattice. It is an algebraic lattice. Its compact elements are the subuniverses $\operatorname{Sg}(X)$ with $X$ finite, that is, the finitely generated ones.

Example: for the group $(\Z, +, -, 0)$ the subuniverses are the sets $n\Z$ with $n \ge 0$. Here $m\Z \subseteq n\Z$ if and only if $n$ divides $m$, and $m\Z \vee n\Z = d\Z$ with $d = \gcd(m, n)$.

Non-example: the lattice of all subsets of $\Z$ is not the subalgebra lattice of this group. The subset $\set{1}$ is not closed under addition, so it is not a subuniverse.

When the signature has no constant symbols, the empty set is a subuniverse and is the least element of $\operatorname{Sub} \mathbf{A}$. Some authors exclude it and call only nonempty subuniverses subalgebras.