Knaster-Tarski Lemma

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

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

Then $f$ has a least fixed point and a greatest fixed point.

Proof
Let $P = \left\{{x \in L: x \preceq f(x) }\right\}$.

Let $p = \bigvee P$, the supremum of $P$.

Let $x \in P$.

Then by the definition of supremum, $x \preceq p$.

Since $f$ is increasing, $f(x) \preceq f(p)$.

By the definition of $P$, $x \preceq f(x)$.

Thus $x \preceq f(p)$ because $\preceq$ is an ordering, and therefore transitive.

As this holds for all $x \in P$, $f(p)$ is an upper bound of $P$.

By the definition of supremum, $p \preceq f(p)$.

Now since $f$ is increasing, $f(p) \preceq f(f(p))$.

Thus $f(p) \in P$ by the definition of $P$.

Since $p$ is the supremum of $P$, $f(p) \preceq p$.

Since we already know that $p \preceq f(p)$, $f(p) = p$ because $\preceq$ is an ordering and therefore antisymmetric.

Thus $p$ is a fixed point of $f$.

Because $\preceq$ is an ordering, and therefore reflexive, every fixed point of $f$ is in $P$, so $p$ is the greatest fixed point of $f$.

Now note that $f$ is also increasing in the dual ordering, so it also has a greatest fixed point in the dual ordering. That is, it has a least fixed point in the original ordering.