I have been reading a book, "Introduction to Lattices and Order", and I'm trying to solve exercise 8.29 as the following in it:
Suppose that $P$ is a complete lattice and let $F$ and $G$ be order-preserving maps. If $F$ and $G$ have a common fixed-point, then they have a least common fixed-point given by $a := \wedge \{x \in P \mid F(x) \le x \mathop{\&} G(x) \le x \}$.
My proof is: for all $y \in \{x \in P \mid F(x) \le x \mathop{\&} G(x) \le x \}$, I have $a \le y$, so $F(a) \le F(y) \le y$ and $G(a) \le G(y) \le y$. Therefore I have $F(a) \le a$ and $G(a) \le a$. Here I want to prove that $a \le F(a)$ and $a \le G(a)$, but it seems hard to do so because I don't assume that F and G are commute.
Could you advise me about this problem?
This bears no resemblance to the first version of this answer.
I’m going to assume that you’ve already seen the Knaster-Tarski theorem, since it seems unlikely that that exercise would precede it. The map
$$\varphi:P\times P\to P\times P:\langle x,y\rangle\mapsto\langle F(x),G(y)\rangle$$
is order-preserving, and $P\times P$ is a complete lattice with the usual product order, so by the Knaster-Tarski theorem the set $M$ of fixed points of $\varphi$ is a complete lattice. Let $$\Delta=\{\langle x,x\rangle\in P\times P:x\in P\}\;.$$ Then $M\cap\Delta$ is the set of common fixed points of $F$ and $G$, so by hypothesis $M\cap\Delta\ne\varnothing$. Thus $M\cap\Delta$ has a least element in $M$, which is what we wished to prove.