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Let $X$ be a complete lattice, and $g$ a function from $X$ to $X$ s.t. $x_1\le x_2$ $\implies g(x_1)\le g(x_2)$. Show that there must be some element in $X$ that maps to itself.

Here is what I am thinking so far. We can take the set $Y=\{x\in X \ | \ x\le g(x)\}$ and from every picture I draw I get that the sup of this set is the element I'm looking for. But I am not quite sure how to prove it. Any ideas would be good.

-thanks

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Hint. Let $a=\sup Y$. Prove the following statements:

  1. $x\in Y\Longrightarrow g(x)\in Y$.

  2. $g(a)$ is an upper bound for $Y$.

  3. $a\in Y$.

  4. $g(a)=a$.

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