6 typo

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.

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.

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 $(-\infty,x_1]$, $(x_n,x_{n+1}]$, $[y_1,y_\infty)$, [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$x_n\to z$. 5 added details; added 2 characters in body; added 8 characters in body 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. 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$(-\infty,x_1]$,$(x_n,x_{n+1}]$,$[y_1,y_\infty)$,$[y_{n+1},y_n)$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$x_n\to z$. 4 added 9 characters in body 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. Proof: Passing to a subsequence, 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 subsequence, we can suppose that$x_{n+2}-x_{n+1}\le\frac12(x_{n+1}-x_n)$and that$x_{n+1}-x_n\le2^{-n-1}$. 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 cover $R$ 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.

3 edited body
2 fixed proof; deleted 79 characters in body; added 14 characters in body
1