Suppose $R$ is an ordered field. Call a continuous map $f: R \rightarrow R$ a contraction if there exists $r < 1$ (in $R$) such that $|f(x)-f(y)| \leq r |x-y|$ for all $x,y \in R$ (where $|x| := \max(x,-x)$). Suppose that every contraction from $R$ to $R$ has a unique fixed point. Must $R$ be the field of real numbers?


For a related question, see [Converse to Banach's fixed point theorem?](https://mathoverflow.net/questions/26119/converse-to-banachs-fixed-point-theorem) .


Jacek Jachymski's article "A discrete fixed point theorem of Eilenberg as a particular case of the contraction principle" ( <http://emis.impa.br/EMIS/journals/HOA/FPTA/2004/131.pdf> ) and the references it contains may be relevant. However, the non-Archimedean metric spaces that the article considers are bounded, which non-Archimedean ordered fields certainly are not. Also, my question is not about metric spaces, since my notion of distance lives in $R$ itself, not the real numbers.



Converse to Banach’s fixed point theorem for ordered fields?
Yes, it is true that $R$ must be the field of real numbers.


As $R$ is an ordered field, it is naturally an extension $\mathbb{Q}\hookrightarrow R$. We can prove the following two properties, which characterize the reals among the ordered fields.



> 
> 1) $\mathbb{Q}$ has no upper bound in $R$ (i.e., $R$ is Archimedean).
> 
> 
> 


*Proof:* Call element $x$ of $R$ *infinite* if $\vert x\vert$ is an upper bound for $\mathbb{Q}$, and *finite* otherwise. Then we can define $f\colon R\to R$ by
$$
f(x)=\begin{cases}
\frac{x}{2}+\frac12\max(x,0)+(2+\max(x,0))^{-1},&\textrm{if }x\textrm{ is finite},\\\\
x/2,&\textrm{if }x\textrm{ is infinite}.
\end{cases}
$$
So,


* If $x,y$ are finite then they have an upper bound $a\ge0$ in $\mathbb{Q}$, and it can be seen that $\vert f(x)-f(y)\vert\le(1-(2+a)^{-2})\vert x-y\vert$.
* If $x,y$ are both infinite then $\vert f(x)-f(y)\vert=\frac12\vert x-y\vert$.
* If $x$ is infinite and $y$ is finite then $\vert f(x)-f(y)\vert\le \frac12\vert x\vert+\vert f(y)\vert\le\frac34\vert x-y\vert$.


In any case, if $\mathbb{Q}$ had an upper bound $\kappa\in R$ then we have $\vert f(x)-f(y)\vert\le (1-\kappa^{-1})\vert x-y\vert$ so that, by hypothesis, $f$ has a fixed point. But it can be seen that $f(x) > x$ for finite $x$ and $f(x)=\frac x2\not=x$ for infinite $x$. So, it doesn't have a fixed point, giving a contradiction.



> 
> 2) Every Cauchy sequence $x\_n$ in $R$ converges.
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*Proof:* Passing to a subsequence1, it can be assumed that $x\_n$ is monotonic, and replacing $x\_n$ by $-x\_n$ if necessary, we can suppose that it is increasing. If it is eventually constant then the result is immediate. Otherwise, by further passing to a subsequence2, we can suppose that $x\_{n+2}-x\_{n+1}\le\frac12(x\_{n+1}-x\_n)$ and that $x\_{n+1}-x\_n < 2^{-n-1}$. Then, $y\_n=x\_n+2^{-n}$ is a strictly decreasing sequence with $0\le y\_n-x\_n\le 2^{-n}$. Again, passing to a subsequence, it can be assumed that $y\_{n+1}-y\_{n+2}\le\frac12(y\_n-y\_{n+1})$.


We can define $f\colon R\to R$ linearly mapping $(-\infty,x\_1]$ onto $(-\infty,x\_2]$, $(x\_n,x\_{n+1}]$ onto $(x\_{n+1},x\_{n+2}]$, $[y\_1,\infty)$ onto $[y\_2,\infty)$, and $[y\_{n+1},y\_n)$ onto $[y\_{n+2},y\_{n+1})$ ($n\ge1$). This can be done such that $\vert f(x)-f(y)\vert\le\frac12\vert x-y\vert$ on each interval, in which case it does not have any fixed points in these intervals. Furthermore, if $x\_n$ had no limit point, then the intervals cover3 $R$ and this defines $f$ everywhere. But, then, $\vert f(x)-f(y)\vert\le\frac12\vert x-y\vert$ for all $x,y\in R$ implying that $f$ has a fixed point, giving a contradiction.




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I'll add a few more details that I passed over rather quickly above. A sequence $x\_n$ is Cauchy if, for each $r > 0$ in $R$ then $\vert x\_n - x\_m\vert < r$ for large enough $m,n$. Any subsequence of a Cauchy sequence is itself Cauchy and tends to a limit $x$ if and only if the orginal sequence tends to $x$.


1 Any sequence in a linearly ordered set has a monotonic subsequence.


2 If $x\_n$ is an increasing Cauchy sequence, which is not eventually constant, then it is possible to choose a subsequence $x\_{n\_k}$ as follows. Once $x\_{n\_k}$ has been chosen, then there is an $m > n\_k$ such that $x\_m \not= x\_{n\_k}$. As it is Cauchy, $n\_{k+1}\ge m$ can be chosen such that $\vert x\_r-x\_s\vert < \min(2^{-k-2},(x\_m-x\_{n\_k})/2)$ for all $r,s\ge n\_{k+1}$. This ensures that $x\_{n\_{k+2}}-x\_{n\_{k+1}}$ is less than both $2^{-k-2}$ and $(x\_{n\_{k+1}}-x\_{n\_k})/2$ for all $k$.


3 If $z\in R$ was not in any of the intervals $(-\infty,x\_1]$, $(x\_n,x\_{n+1}]$, $[y\_1,\infty)$, $[y\_{n+1},y\_n)$ then $x\_n < z < y\_n$ for all $n$. So, $\vert z-x\_n\vert\le y\_n-x\_n\le 2^{-n}$. Given any $r > 0$ in $R$, the fact that we have already shown $R$ to be Archimedean in (1) implies that $2^n > r^{-1}$ for large $n$. So, $\vert z - x\_n\vert < r$ for large $n$, and $x\_n\to z$.


