Tarski's HSP formula

Remarks

Each of $H$, $S$ and $P$ maps $V(K)$ into itself, so $HSP(K) \subseteq V(K)$. It remains to show that $HSP(K)$ is closed under $H$, $S$ and $P$. This follows from the rules $SH \le HS$, $PS \le SP$, $PH \le HP$ and from the idempotence of each operator. For example, $S\,HSP \le HS\,SP = HSP$ and $P\,HSP \le HP\,SP \le HSP\,P = HSP$. Example: for $K = \set{\Z}$ in the signature of groups, $HSP(K)$ is the class of all abelian groups. Indeed, every abelian group is a quotient of a free abelian group, and a free abelian group embeds in a power of $\Z$.

Going deeper

For a student

The formula gives the standard way to find the members of $V(A)$ for one algebra $A$. Each member of $V(A)$ is a homomorphic image of a subalgebra of a power $A^I$. In particular, the free algebras of $V(A)$ are subalgebras of powers of $A$, which is the starting point of the study of free spectra. In the congruence-distributive case, Jónsson's lemma sharpens the formula for the subdirectly irreducible members.