Knaster-Tarski Lemma

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

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:
 * $f \left({x}\right) \preceq f \left({p}\right)$

By the definition of $P$:
 * $x \preceq f \left({x}\right)$

Thus because $\preceq$ is an ordering, and therefore transitive:
 * $x \preceq f \left({p}\right)$

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

By the definition of supremum:
 * $p \preceq f \left({p}\right)$

As $f$ is increasing:
 * $f \left({p}\right) \preceq f \left({f \left({p}\right)}\right)$

Thus by the definition of $P$:
 * $f \left({p}\right) \in P$

Since $p$ is the supremum of $P$:
 * $f \left({p}\right) \preceq p$

Since we already know that $p \preceq f \left({p}\right)$:
 * $f \left({p}\right) = p$

because $\preceq$ is an ordering and therefore antisymmetric.

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

We have that $\preceq$ is an ordering, and therefore reflexive.

Thus 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.

Thus $f$ also has a greatest fixed point in the dual ordering.

That is, it has a least fixed point in the original ordering.