# Using supremum property to find a point such that f(c)=c.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a nondecreasing function.

Let $a<f(a)$ and $f(b)<b$. Prove that there is a $a<c<b$ such that $f(c)=c$.

My attempt at a proof is as follows. Let $c:=\sup\{x:a\leq x\leq b\text{, }x\leq f(x)\}$.

This is where I'm stuck. Since I can't use more powerful theorem such as the IVT I find this problem far more complex.

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If you define $c$ in that way, what you can tell us about $c$, $f(c)$, and other values that aren't $c$? –  Hurkyl Sep 13 '12 at 4:59

HINT: Let $L=\{x\in[a,b]:x<f(x)\}$. Is $L$ non-empty? Is $L$ bounded above?

Added: In other words, your idea is reasonable, though I used a slightly different set from yours. $L$ is non-empty and bounded (why?), so we can let $c=\sup L$. What do you know about $f(x)$ for $x\in[c,b]$? What happens if $f(c)\ne c$? Remember, $f$ is non-decreasing.

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now im interested. you have to use completeness axiom for this, right? –  Taylor Sep 13 '12 at 4:59
@Taylor: Right; in this context that’s the least supremum property. –  Brian M. Scott Sep 13 '12 at 5:03
And it is needed. If we were talking about $\mathbb Q$, we could select a sequence $(a_n)$ converging to $\sqrt 2$ from below and $(b_n)$ from above. Then let $f(x)$ be the minimal $a_n>x$ for $x<\sqrt 2$ and the maximal $b_n<x$ for $x>\sqrt 2$. –  Hagen von Eitzen Sep 13 '12 at 5:23

Define another function $g(x) = f(x) - x$. Then try to find out if this function crosses $0$.

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That won’t help, since the OP isn’t allowed to use the intermediate value theorem (and is in fact apparently expected to use the least upper bound property). –  Brian M. Scott Sep 13 '12 at 4:56
ah you're right. didn't read the last line. sorry about that. –  Taylor Sep 13 '12 at 4:57
Note that $f$ isn't continuous anyways. –  Hurkyl Sep 13 '12 at 4:59