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I need a hint for the following question:

Let $S$ be a nonempty ordered set such that every nonempty subset $E\subseteq S$ has both a least upper bound and a greatest lower bound. Suppose $f:S \rightarrow S$ is a monotonically increasing function. Show that there exists an $x\in S$ so that $f(x)=x$.

My reasoning so far has been as follows:

I wanted to prove this by contradiction. If we assume the statement is false, this implies that for each $x\in S$, $f(x) > x$ or $f(x) < x$. Let $A:=\{x:f(x)<x\}$, $B:=\{x:f(x)>x\}$.

Let's assume $A$ is empty. Then it is easy to show a contradiction. We know that $\sup(B)\in B$. But that means that $f(\sup(B))>\sup(B)$ which is impossible. The same idea applies if we assume $B$ is empty.

What I am trying to do is prove the case where both $A$ and $B$ are nonempty. To prove the first part I haven't even used the monoticity of the function $f$, so I know that I have to use it at this point. That means trying to find a contradiction by finding $x,y\in S$, $x\le y$, $f(x)>f(y)$. I was trying to look at the supremum and infimum of $A$ and $B$ but that didn't seem to lead me anywhere.

I would appreciate hints!

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This is a tricky pure real analysis problem. Drawing functions helps very much.

You only need to consider the case $B \neq \{\}$

Let $b= \sup B$

Take a look at the red point below : whatever the function, it's a fixed point. And it's also $\sup B$ on the $x$ axis.

enter image description here

  • if $b\in B$, then $f(b)>b$. Applying $f$, $f(f(b))>f(b)$

Hence $f(b) \in B$.

But by $b=\sup B$ definition as an upper bound of $B$, $f(b) \leq b$

Since $f(b)>b$, it's a contradiction.

  • if $b\notin B$ then $f(b)\leq b$ .

Consider $x\in B$

Then since $b\notin B$ and $b=\sup B$ (in other words, $b$ is an upper bound of$B$ that does not belong to $B$), it holds $x<b$

Applying $f$, $f(x)<f(b)$

Since $x\in B$, $x<f(x)$, hence $x<f(x)<f(b)$

Hence $x<f(b)$

This proves that $f(b)$ is an upper bound of $B$.

Since $b=\sup B$, $f(b) \geq b$ (using the least upper bound definition)

We have that $f(b) \geq b$ and $f(b) \leq b$, hence $f(b)=b$

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  • $\begingroup$ Nice.. I am upset I didn't go far enough with "using" the function twice..Thanks a lot! $\endgroup$ – daniel May 12 '14 at 10:29
  • $\begingroup$ Now, you can try to solve this one: >Let $f,g:\mathbb [a,b] \to \mathbb [a,b]$ be monotonically increasing functions such that $f\circ g=g\circ f$. Prove that $f$ and $g$ have a common fixed point. $\endgroup$ – Gabriel Romon May 12 '14 at 10:41
  • $\begingroup$ Alright then, I'm on it :) $\endgroup$ – daniel May 12 '14 at 11:01
  • $\begingroup$ It seems that you are assuming that $f$ is strictly increasing (correct me if I am wrong). Unless I am mistaken, the same proof works for weakly increasing functions if you define $B = \{x:f(x) \ge x\}$. $\endgroup$ – Martin R Apr 26 '17 at 9:53

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