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I am wondering how to show:
An order-preserving map $f$ of a complete lattice $A$ into itself has at least one fixed element.

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Let $B=\{a\in A:a\le f(a)\}$; $B\ne\varnothing$, since $0_A\in B$. Let $a\in B$; then $f(a)\le f\big(f(a)\big)$, since $f$ is order-preserving. $A$ is complete, so let $u=\sup B$; I’ll leave it to you to finish by showing that $f(u)=u$.

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    $\begingroup$ If $u>f(u)$, then $\exists a\in B$ s.t. $ f(u) < a \le u$. So $f(f(u))\le f(a)\le f(u)$. But then $a>f(u)\ge f(a)$, a contradiction. So $u\in B$! $\endgroup$ – Allitee Jan 17 '16 at 16:53
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    $\begingroup$ @Allitee: Almost, but remember, $u\not\le f(u)$ does not necessarily mean that $u>f(u)$: $u$ and $f(u)$ might be incomparable, so you have to work a little harder. For each $a\in B$ we know that $a\le u$, so $a\le f(a)\le f(u)$; this shows that $f(u)$ is an upper bound for $B$. Now we can conclude that $u\le f(u)$, since $u$ is the least upper bound for $B$. But then $f(u)\le f\big(f(u)\big)$, so $f(u)\in B$, and therefore $f(u)\le u$. Hence $f(u)=u$. $\endgroup$ – Brian M. Scott Jan 17 '16 at 22:25

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