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 (or class) of fixed points of $f$.

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

Proof
Let $S \subseteq F$.

Let $s = \bigvee S$ be the supremum of $S$.

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 the upper closure, $U$, of $s$.

Note that $U = \left[{{s}\,.\,.\,{\top}}\right]$, the closed interval between $s$ and the top element of $L$.

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 Interval in Complete Lattice is Complete Lattice, $(U, \preceq)$ is a complete lattice.

Thus by Knaster-Tarski Lemma, $f$ has a least fixed point in $U$.

Thus $S$ has a supremum in $F$.

A precisely similar argument shows that $S$ has an infimum in $F$.

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

Also see

 * Knaster-Tarski Lemma
 * Knaster-Tarski Lemma: Power Set form

Tarski, however, appears to claim sole credit.