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The Knaster–Tarski fixed-point theorem states that if $L$ is a complete lattice and $f:L\to L$ is order-preserving, then the fixed points of $f$ form a complete lattice; in particular, $f$ has least and greatest fixed points.

LetFor every $x\in P$, monotonicity gives $f(a)\le f(x)\le x$, so $f(a)\le a$. Applying $f$ once more shows $f(f(a))\le f(a)$, hence $f(a)\in P$. Therefore $a\le f(a)$ by the definition of $a$, and $f(a)=a$. Every fixed point belongs to $P$, so $a$ is the least fixed point. Dually, $\bigvee \{x:x\le f(x)\}$ is the greatest fixed point. For any family $S$ of fixed points, apply the same argument inside the upper interval above $\bigvee S$ to obtain the least fixed point above every member of $S$; this is their join in the fixed-point set. The dual construction supplies meets, proving the full statement.

To deduce the Cantor-Schröder-Bernstein theorem, let $i:A\to B$ and $j:B\to A$ be injections. On the complete lattice $\mathcal{P}(A)$ defineThis map is order-preserving, so it has a fixed point $X$. The relationshows thatis a bijection: its two pieces map bijectively onto the disjoint sets $i(X)$ and $B\setminus i(X)$. Thus injections both ways imply a bijection.

Let $P$ be the poset of countable subsets of $\mathbb{R}$. The family of all singletons $\{\{r\}:r\in \mathbb{R}\}$ has no upper bound in $P$, since any upper bound would contain every real number and would be uncountable. Therefore

Finally, use the well-ordering theorem to fix a well-order $\prec$ of $\mathbb{R}$. Every countable $A$ omits some real number; let $m(A)$ be its $\prec$-least omitted element and defineIf $A\subseteq B$, then either $m(A)\in B$, or every point preceding $m(A)$ lies in $A\subseteq B$ and $m(A)=m(B)$. In either case $f(A)\subseteq f(B)$, so $f$ is order-preserving. Yet $m(A)\notin A$, and henceThis does not contradict Knaster–Tarski because $P$ is not a complete lattice.