# Knaster-Tarski Lemma/Corollary/Power Set/Proof 2

## Theorem

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 fixed point.

## Proof

Let $P = \set {x \in \powerset S: x \subseteq \map f x}$.

Let $\ds \bigcup P$ be the union of $P$.

Let $x \in P$.

$\ds x \subseteq \bigcup P$

Since $f$ is increasing, $\map f x \subseteq \map f {\bigcup P}$.

By definition of $P$, $x \subseteq \map f x$.

Thus $\ds x \subseteq \map f {\bigcup P}$ by Subset Relation is Transitive:

As this holds for all $x \in P$, by Union is Smallest Superset: General Result:

$\ds \bigcup P \subseteq \map f {\bigcup P}$

Now since $f$ is increasing:

$\ds \map f {\bigcup P} \subseteq \map f {\map f {\bigcup P} }$

Thus $\ds \map f {\bigcup P} \in P$ by definition of $P$.

Therefore by Set is Subset of Union: General Result:

$\ds \map f {\bigcup P} \subseteq \bigcup P$

We have that:

$\ds \bigcup P \subseteq \map f {\bigcup P}$

Thus by definition of set equality:

$\ds \map f {\bigcup P} = \bigcup P$

Thus $\ds \bigcup P$ is a fixed point of $f$.

$\blacksquare$

## Source of Name

This entry was named for BronisÅ‚aw Knaster and Alfred Tarski.