Apparent counterexample in the theorem that term functions form a clone.

Question

Consider the structure $(\mathbb{R};)$, that is, just the set $\mathbb{R}$ with no operations, constants, or relations. We have a countably infinite set of variables $x_1, x_2, x_3, ...$. A term function, in this structure, is simply represented by a variable, since there are no operations or constants. I have read that term functions always form a clone. In this case, it would be the minimal clone of all and only the projection functions. However, I don't think we actually get all the projection functions. For example, consider the 3-ary projection function which gives the second coordinate. I don't think there is a term function that represents the function. Because, for any $x_n$, it represents the n-ary projection function that gives the n-th coordinate. I am puzzled by this. I thought term functions always form a clone. I would like a clarification of what a term function actually is, and the proof that term functions, under a suitable definition, actually form a clone.

Answer 1

I follow the terminology in Burris and Sankappanavar, on the definition of projections.
There,

Definition II.10.2. Given a term $p(x_1, \ldots, x_n)$ of type $\mathscr F$ over some set $X$ and given an algebra $\mathbf A$ of type $\mathscr F$ we define a mapping $p^{\mathbf A}:A^n \to A$ as follows:

(1) if $p$ is a variable $x_i$, then $$p^{\mathbf A}(a_1,\ldots,a_n)=a_i$$ for $a_1, \ldots, a_n \in A$, i.e., $p^{\mathbf A}$ is the $i$th projection map;
(2) [...]

So $x_n$ represents any projection function $p$ on $m \geq n$ coordinates such that $p(a_1, \ldots, a_m) = a_n$, and not just the $n$-ary projection.

That must be the source of your confusion.
It's possible that other authors have a different definition, but I suppose it must be equivalent.
So in your particular case $p(x_1,x_2,x_3)=x_2$ is given precisely by the variable $x_2$.