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I got this problem:

Prove that if $f:[a,b]\to[a,b]$ is a nondecreasing function then $\exists x_0\in[a,b]$ such that $f(x_0)=x_0$ (i.e. $f$ has a fixed point).

(Hint: set $A=\{x\in[a,b]|x\leq f(x)\}$ and show that $x_0=\sup f([a,b])$ exist and that $f(x_0)=x_0$)

I tried to show that $A\neq\emptyset$ by supposing that $A=\emptyset$ and trying to reach a contradiction, but I got stuck.

Thanks only help.

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    $\begingroup$ Note that $a \in A$, so $A \neq \emptyset$. $\endgroup$ – Crostul Oct 4 '14 at 8:24
  • $\begingroup$ Thanks, I totally missed it. $\endgroup$ – MathNerd Oct 4 '14 at 8:30
  • $\begingroup$ You've got your hint wrong. It should read $x_0 = \sup f(A)$. $\endgroup$ – posilon Oct 4 '14 at 9:51
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Hints:

$A\not =\emptyset.$ For example, $a\in A$ since $f(a)\in [a,b]$, then $a \le f(a)$.

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  • $\begingroup$ You are welcome. $\endgroup$ – Paul Oct 4 '14 at 8:31

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