Idempotency and Fixed-point combinators In the $\lambda$-calculus, for a fixed-point combinator $P$, we have $Pf = f(Pf)$ for all functions $f$. Thus we could always expand $(Pf)$ as follows:
$$Pf = f(Pf) = f(f(Pf)) = f(f(f(Pf))) = f(f(f(…f(Pf)…))) = f^{\circ n}(Pf)$$
where $f^{\circ n}$ is the $n$-fold composition of $f$.
In some sense, it seems like $Pf$ is the $\infty$-fold composition of $f$. Or $Pf = f^{\circ n}(Pf)$ for arbitrarily large $n$.
We have $Pf = f(Pf) = f(f(Pf)) = f(f(f(Pf))) = \dots$, meaning that

*

*$Pf$

*$f(Pf)$

*$f(f(Pf))$

*$f(f(f(Pf)))$

*etc.

are all the same. What does that say about $f$? Does that imply that $f$ is idempotent?
 A: A function $f \colon S \to S$ is idempotent if $f \circ f = f$, that is, if for every $x \in S$ one has $f(f(x)) = f(x)$.
A function $f \colon S \to S$ has a fix point if there is some $x \in S$ such that $f(x) = x$ (and such a $x$ is called a fix point of $f$).
Therefore, idempotence is a global property of a function (it says something about all the elements in its domain), while having a fix point is a local property of a function (it claims the existence of some specific element in its domain).
The fact that a function has a fix point or even infinitely many fix points, does not imply that the function is idempotent. Take for instance the function $x \mapsto x\sin x$, whose fix points are $\{\frac{\pi}{2} + 2k\pi \mid k \in \mathbb{Z}\}$, but clearly it is not idempotent.
The same holds for the $\lambda$-calculus. From the fact that every $\lambda$-term $f$ has a fix point (which is $Pf$, where $P$ is a fixpoint combinator), or even  infinitely many fix points (it can be proved), it does not follow that $f$ is idempotent. Indeed, it would imply that $\lambda$-terms can only represent idempotent functions. But it is well known that the $\lambda$-calculus is Turing-complete and hence it can represent all recursive functions, including the non-idempotent ones.
