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

Theorem
Let $S$ be a set.

Let $\mathcal P \left({S}\right)$ be the power set of $S$.

Let $f: \mathcal P \left({S}\right) \to \mathcal P \left({S}\right)$ be a $\subseteq$-increasing mapping.

That is, suppose that for all $T, U \in \mathcal P \left({S}\right)$:


 * $T \subseteq U \implies f \left({T}\right) \subseteq f\left({U}\right)$

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

Proof
Let $L$ be the power set of $S$.

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

Let $p = \bigcup P$, the union of $P$.

Let $x \in P$.

Then by Subset of Union, $x \subseteq p$.

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

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

Thus $x \subseteq f(p)$ because $\subseteq$ is transitive.

As this holds for all $x \in P$, $p \subseteq f(p)$ by Union Smallest.

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

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

Since $p$ is the union of $P$, $f(p) \subseteq p$.

Since we already know that $p \subseteq f(p)$, $f(p) = p$ by Equality of Sets.

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

Because every set is a subset of itself, every fixed point of $f$ is in $P$, so $p$ is the greatest fixed point of $f$.

The argument can be swapped around, using intersections instead of unions, to show that $f$ has a least fixed point as well.