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Let $\mathbb{F} $ be a ordered field and a function $f:\mathbb{F}\to \mathbb{F}$ is called monotonically increasing if $\forall x,y \in \mathbb{F} $: $ x \leq y \Rightarrow f(x) \leq f(y)$. Let $ a,b \in \mathbb{F} $: and $a <b$ and $f:\mathbb{F}\to \mathbb{F}$ a monotonically increasing function with $f(a) > a$ and $f(b) < b$.

Is there for 1) $\mathbb{F} = \mathbb{R } $ 2) $\mathbb{F} = \mathbb{Q }$ a point with $f(x) = x$ ?

Until now we have : Infimum and supremum, axiom of real numbers,intervals etc. but no theorems about monotonically functions

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  • $\begingroup$ For $\Bbb R$, yes. For $\Bbb Q$, no. $\endgroup$ – Clayton Oct 17 '17 at 19:23
  • $\begingroup$ For $\Bbb Q$, consider $f(x)=x^3-\frac1{16}$ with $a=-\frac12$, $b=0$ $\endgroup$ – Hagen von Eitzen Oct 17 '17 at 19:29
  • $\begingroup$ Restrict the function from f(a) to f(b) and apply brouwer fixed point. This should work right?! $\endgroup$ – Shreedhar Bhat Oct 17 '17 at 19:30
  • $\begingroup$ Yes that was also my intuition, because Q is a dense set in R. But i cannot finde a counterexample for Q or prove my intuition for R $\endgroup$ – Anna Saabel Oct 17 '17 at 19:30
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    $\begingroup$ @ChangWoo The "obvious" idea is to pick a function such that the wanna-be fixpoint is irrational, e.g., such that it must be a root $\pm\sqrt 2$ of the polynomial $x^2-2$. Now turn this condition into an increasing function (and I think I did that wrong in my suggestion): We want $f(x)=x\implies x^2=2$, so we might try $f(x)=x+x^2-2$, but that is not increasing for $x\ll 0$, and also to find $a$ and $b$ it should be increasing slower that the identity function (at least near the fixpoint). So next try is $f(x)=x+2-x^2$ which gives us $f(a)>a$ for $a=1$ and $f(b)<b$ for $b=2$ (continued) $\endgroup$ – Hagen von Eitzen Oct 18 '17 at 6:08
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For $\Bbb F=\Bbb R$, let $a_0=a$, $b_0=b$, and define $a_n,b_n$ recursively: If $f(\frac{a_n+b_n}2)>\frac{a_n+b_n}2$, let $a_{n+1}=\frac{a_n+b_n}2$, $b_{n+1}=b_n$; if $f(\frac{a_n+b_n}2)<\frac{a_n+b_n}2$, let $a_{n+1}=a_n$, $b_{n+1}=\frac{a_n+b_n}2$ (and if $f(\frac{a_n+b_n}2)=\frac{a_n+b_n}2$, we are done). Show that $a_n<f(a_n)\le f(b_n)<b_n$ for all $n$, and that $a_n,b_n$ converge to some $c\in\Bbb R$. Conclude that $f(c)=c$.

For $\Bbb F=\Bbb Q$, consider $$f(x)=\begin{cases}x+\frac14(2-x^2)&x\le 2\\x-\frac12&x>2\end{cases} $$ and verify that it is increasing and fixpoint-free, and that with $a=1$, $b=2$, we have $a<f(a)<f(b)<b$ as desired.

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