Tarski’s fixed-point theorem and the existence of minimal fixed points

Question

Let $(X,\geq)$ be a partially ordered set. Suppose that

• any non-empty subset of $X$ has a supremum in $X$ with respect to the order $\geq$;
• $X$ has a maximum element: $\exists \overline x\in X:\overline x\geq x\,\,\forall x\in X$;
• $X$ has a minimum element: $\exists \underline x\in X:x\geq\underline x\,\,\forall x\in X$.

Let $f:X\to X$ be a non-decreasing function; that is, for any $x,y\in X$, $x\geq y$ implies that $f(x)\geq f(y)$.

Let $\mathscr F\equiv\{x\in X\,|\,x=f(x)\}$ be the set of fixed points of $f$.

Now, Tarski’s fixed-point theorem implies that $\mathscr F$ is not empty. It is also easily proved that $\mathscr F$ has a maximum element.

My question is: does $\mathscr F$ necessarily admit a minimum element as well?

Note that the existence of infima of non-empty subsets of $X$ has not been assumed, unlike the existence of suprema.

Any hints, either towards a proof or towards a counterexample, would be greatly appreciated.

Answer 1

The existence of infima does not have to be assumed, it follows from the existence of suprema. Consider any set $A\subseteq X.$ Let $B$ be the set of all lower bounds for $A.$ Then $B\ne\emptyset$ since $X$ has a minimum element. It is easy to see that the supremum of $B$ is also the infimum of $A.$ So the answer to your question is yes, $\mathscr F$ has both a maximum and a minimum element.