I am trying to show the claim in the title. I have made some progress, which I will outline below:
Denote the complete lattice $(X, \leq)$, and an arbitrary order endomorphism on it $\phi$. I propose that, given the set $A = \{ x \in X \mid x \leq \phi(x) \}$ (whose existence is assured by $\bot X$), $\vee A = \phi(\vee A)$ (the join operation being acceptable because $X$ is complete). To this end, I want to show:
- $\vee A \leq \phi(\vee A)$
- $\phi(\vee A) \leq \vee A$.
Unfortunately, I got lost very quickly after that. I have noted a few relationships but that's about as far as I've gotten. (I use $a$ to represent an arbitrary member of $A$.)
- $a \leq \vee A$ by definition of join
- $\phi(a) \leq \phi(\vee A)$ by (1) and the order preserving nature of $\phi$
- $a \leq \phi(a)$ by construction of $A$
- $\phi(a) \leq \vee \phi[A]$ by definition of join
- $\phi(\vee A) = \vee \phi[A]$ by... I'm actually not sure about this, haha. Might need to prove this as a lemma, if I end up using it.
I'd be very grateful if anyone could help me with this proof. Please let me know if you need anything more from me.
Let $u=\bigvee A$. Then $x\le u$ for each $x\in A$, so $x\le\varphi(x)\le\varphi(u)$, and it follows that $\varphi(u)$ is an upper bound for $A$ and hence that $u\le\varphi(u)$ (since $u$ is the least upper bound of $A$). But then $\varphi(u)\le\varphi\big(\varphi(u)\big)$, so $\varphi(u)\in A$. Can you finish it from here?