Any order preserving map of a poset into itself has a fixed point

Question

In the book of Algebra by Hungerford, at pafe 15, it is claimed that

However, consider $(\mathbb{R}, \leq)$ and $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x)=x+1$. It is a complete partially ordered set with an order preserving map. However, this map do not have any fixed point, so am I missing something in here ?