Knaster-Tarski Lemma

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Theorem

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

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


Then $f$ has a smallest 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 $P = \set {x \in L: x \preceq \map f x}$.

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:

$\map f x \preceq \map f p$

By the definition of $P$:

$x \preceq \map f x$

Thus because $\preceq$ is an ordering, and therefore transitive:

$x \preceq \map f p$

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


By the definition of supremum:

$p \preceq \map f p$

As $f$ is increasing:

$\map f p \preceq \map f {\map f p}$

Thus by the definition of $P$:

$\map f p \in P$

Since $p$ is the supremum of $P$:

$\map f p \preceq p$

Since we already know that $p \preceq \map f p$:

$\map f p = p$

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

Thus $p$ is a fixed p oint 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.


Sources

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