Knaster-Tarski Theorem

Theorem
Let $(L, \preceq)$ be a complete lattice.

Let $f: L \to L$ be an increasing mapping.

Let $F$ be the set of fixed points of $f$.

Then $(F, \preceq)$ is a complete lattice.

Proof
Let $S \subseteq F$.

Let $s = \bigvee S$.

We wish to show that there is an element of $F$ that succeeds all elements of $S$ and is the smallest element of $F$ to do so.

By the definition of supremum, an element succeeds all elements of $S$ iff it succeeds $s$.

Thus we seek the least fixed point of $f$ that lies in $U = {\uparrow} s$

First we show that $U$ is closed under $f$.

For any $a \in S$, $a \preceq s$, so $a = f(a) \preceq f(s)$.

Thus $f(s)$ is an upper bound of $S$, so by the definition of supremum, $s \preceq f(s)$.

Let $x \in U$.

Then $s \preceq x$, so $f(s) \preceq f(x)$.

Since $s \preceq f(s)$, $s \preceq f(x)$, so $f(x) \in U$.

Thus the restriction of $f$ to $U$ is an increasing mapping from $U$ to $U$.

By User:Dfeuer/Upper Closure of Element of Complete Lattice is Complete Lattice, $(U, \preceq)$ is a complete lattice.

Thus by User:Dfeuer/Knaster-Tarski Theorem/Lemma, $f$ has a least fixed point in $U$.

Thus $S$ has a least upper bound in $F$.

Since this holds for all $S \subseteq F$, $(F, \preceq)$ is a complete lattice.

Remark
The power set form of the Knaster-Tarski Lemma was discovered by Knaster and Tarski.

This theorem was actually by Tarski alone, but it has come to be known as the Knaster-Tarski Theorem.