# Are there infinitely many clones containing all the constant functions?

A clone on a set $$A$$ is a collection of $$m$$-ary functions $$A^m \to A$$ ($$m$$ depends on the function) that is closed under composition and includes all projection functions. Clones are significant because for any language of possible operations on $$A$$, the functions definable from those operations is a clone. Let $$n = |A|$$.

Is it true for all $$n \in \mathbb{N}$$ that there are only finitely many clones on $$A$$ containing all constant functions? That is, for each $$a \in A$$ the clone contains a function $$f: A \to A$$, such that $$f(x) = a$$ for all $$x$$.

### Motivation

The classic case is where $$A = \{0,1\}$$; a set of functions is "expressively adequate" if it generates the clone of all functions. It is a famous and fundamental result that sets such as $$\{\lor, \lnot\}$$ and $$\{\to, \bot\}$$ are expressively adequate. A more powerful and incredibly beautiful result is Post's lattice which completely characterizes ALL countably many possible clones on $$\{0,1\}$$. Wikipedia states: "the lattice of clones on a three-element (or larger) set [has] the cardinality of the continuum, and a complicated inner structure."

But I have an alternate definition of expressive adequacy in mind where we wish to always allow constants $$0$$ and $$1$$. Thus we restrict to those clones containing the constant functions, and we find that there are only finitely many clones on $$\{0,1\}$$. In fact, there are 7 clones:

• $$UM$$, which contains just constant functions and projections
• $$\Lambda$$, the set of conjunctive functions
• $$V$$, the set of disjunctive functions
• $$U$$, the set of essentially unary functions (depend only one one coordinate)
• $$A$$, the set of affine functions
• $$M$$, the set of monotone functions
• $$\top$$, all functions.

Thus I am interested in whether there are still finitely many such clones for a set of size greater than $$2$$.