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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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I really think you ought to accept Asaf's answer instead. – Henning Makholm Mar 23 '14 at 15:03
@Henning: I'm starting to think it's not gonna happen. :P – Asaf Karagila Mar 22 '15 at 17:13

Simply note that monotonicity in this context means $A\subseteq B\implies F(A)\subseteq F(B)$.

Now consider the set $D=\bigcup\{A\mid A\subseteq F(A)\}$. Show that $D\subseteq F(D)$ and conclude from that $F(D)\subseteq D$.

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How we're going to prove that without using the term (complete lattice)?!

Define a ruskomsnusk to mean a partially ordered set such that every subset has a least upper bound and a greatest lower bound.

Then $\mathcal P(X)$ ordered by set inclusion forms a ruskomsnusk.

et cetera

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I LOL'd. $\stackrel{\bullet.\odot}{\large\smile}$ – Asaf Karagila Mar 23 '14 at 12:30
Ruskomsnusk must be delicious. – user132181 Mar 23 '14 at 12:47

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