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Let be $F$ be a compact subset of $\mathbb{R}$ and $f: F\to F$ a continuous and non-decreasing function. Prove that $f$ has an fixed point.

I've done the connected case (using $g(x)=f(x)-x$ and the mean value theorem), but in the case that $F$ is not connected, the result still valid?

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Original question was with $F$ closed, not compact: The claim in that generality is false. Let $F=\mathbb R$ and consider $f:F\to F$ given by $f(x)=x+1$. Then $F$ is closed, $f$ is continuous and non-decreasing, but of course can't have a fixed point. Notice that $F$ is also connected, so you cannot have possibly proven the result for the connected case either.

Question was changed to $F$ being compact: Now the claim is true. Here is a proof. Let $S=\{x\in F\mid f(x)<x\}$. If $S$ is nonempty then it has a supremum, $s=\sup (S)$. Let $x_n$ be a sequence in $S$ converging to $s$. Then by continuity of $f$ it follows that $f(s)\le s$. If $f(s)<s$ then for all $x\in S$ holds that $f(x)<s$ (since otherwise it would follow from $x\le s$ and from monotonicity of $f$ that $s=f(x)\le f(s)<s$, a contradiction). Now define a sequence $y_1=s$, and $y_{k+1}=f(y_k)$. Then it follows that $(y_n)$ is monotonically non-increasing and bounded so it converges to some $y$. Continuity of $f$ now implies immediately that $f(y)=y$.

This completes the proof in the case where the set $S=\{x\in F\mid f(x)<x\}$ is not empty. If it is empty that repeat the arguments above for the set $S=\{x\in F\mid f(x)>x\}$, if that set is not empty. If both are empty then $f$ must be the constant function and thus have (lots and lots) of fixed points.

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  • $\begingroup$ The question was changed to restrict to compact $F$. $\endgroup$ – Brandon Carter Mar 6 '13 at 1:30
  • $\begingroup$ I edited the question,$F$ must be limited. I'm sorry for my mistake. $\endgroup$ – Jön Mar 6 '13 at 1:31
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Hint: consider the limit of a sequence $x_n$ with $x_0 \in F$ and $x_{n+1} = f(x_n)$.

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