Assume that $F: \mathcal{P}A \to \mathcal{P}A$ and that $F$ has the monotonicity property: $$X \subseteq Y \subseteq A \implies F(X) \subseteq F(Y)$$ Define $$B = \bigcap \{ X \subseteq A | F(X) \subseteq X \} \; \mathtt{and} \; C = \bigcup \{ X \subseteq A | X \subseteq F(X) \}$$ Show that $F(B)=B$ and $F(C)=C$.      So far I can get: 1. $F(A) \subseteq A \;\&amp;  \; \varnothing \subseteq F(\varnothing)$ 2. given any $X_1 \subseteq F(X_1)$ and $F(X_2) \subseteq X_2$ $(X_1 ,X_2 \subseteq A)$     I have  $B \subseteq X_2 ,F(B) \subseteq F(X_2) \subseteq X_2$ and  $X_1 \subseteq C, X_1 \subseteq F(X_1) \subseteq F(C)$    But: I cannot find a way to show the relation between $F(B)$ and $B$.What should I do?
Let $P(X)$ denote the power set of $X$ (the set of all subsets of $X$) with a partial order given by inclusion.  If $F: P(X) \to  P(X)$ is monotone (order preserving), then $F$ has a fixed point.  How are we going to prove that without using the term (complete lattice)?!
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