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.

  • 1
    $\begingroup$ 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


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.