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.

Power Set

Let $S$ be a set.

Let $\powerset S$ be the power set of $S$.

Let $f: \powerset S \to \powerset S$ be a $\subseteq$-increasing mapping.

That is, suppose that for all $T, U \in \powerset S$:

$T \subseteq U \implies \map f T \subseteq \map f U$

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

Corollary

Let $\struct {L, \preceq}$ be a complete lattice.

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

Then $f$ has a 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.

$\blacksquare$

Source of Name

This entry was named for Bronisław Knaster and Alfred Tarski.