Complete Lattice and fixed point

Question

I am wondering how to show:
An order-preserving map $f$ of a complete lattice $A$ into itself has at least one fixed element.

Answer 1

Let $B=\{a\in A:a\le f(a)\}$; $B\ne\varnothing$, since $0_A\in B$. Let $a\in B$; then $f(a)\le f\big(f(a)\big)$, since $f$ is order-preserving. $A$ is complete, so let $u=\sup B$; I’ll leave it to you to finish by showing that $f(u)=u$.