I am having trouble understanding the definition of a contractive function. The definition is:
Let $(\Bbb R,d)$ be a Metric Space with the standard metric.
Let $A\subseteq X$ (with the same metric as $\Bbb R$)
Let $\,f:A\rightarrow A$ be a function.
$f$ is said to be a contraction on $A$ if there exists some $C\in (0,1)$ such that $\forall x,y\in A$
$d(f(x),f(y))\le C*d(x,y)$
Let $\epsilon>0$
Let $A=[1+\epsilon, + \infty]$
I think $f(x)=\ln(x)+1$ is contractive on $A$ because $\cfrac{d(f(x),f(y))}{d(x,y)}<1,\,\,\forall x,y,\in A$ but at no point is there a fixed point $(f(x)=x)$ $(x\in A$) so it should not be contractive by contrapositive
(I).If $\epsilon >0$ and $A\subset (1,\infty)$ with $\inf A=1+\epsilon$ then $f$ is not contractive on $A$ because $f$ does not map $A$ into $A.$
For $y>0$ we have $\log (1+y)<y$ so $f(1+y)<1+y$ . So $f(\inf A)<\inf A.$ Let $v=\frac {1}{2} (\inf A-f(\inf A)).$
Now $f$ is continuous on $\Bbb R^+$ so there exists $u\in (0,\inf A) $ such that $|x-\inf A|<u\implies |f(x)-f(\inf A)|<v.$ Then $|x-\inf A|<u\implies f(x)<\inf A.$
By definition of $\inf A,$ there exists $x\in A$ with $\inf A\leq x<u+\inf A,$ and for such $x$ we have $f(x)<\inf A,$ so $x\in A$ but $f(x)\not \in A.$
In particular , $f$ is not a contraction on $A=[1+\epsilon,\infty)$ for any $\epsilon >0$ because $f$ does not map $A$ into $A.$
(II).$f$ is not contractive on $[1,\infty).$ For $1<y$ there exists $z\in (1,y)$ such that $$\frac {d(f(y),f(1))}{d(y,1)}= \frac {f(y)-f(1)}{y-1} = f'(z)=1/z>1/y .$$ $$\text {Therefore }\quad \sup_{1<x< y} \frac {d(f(y),f(x))}{d(y,x)}\geq \sup_{1<y}\frac {d(f(y),f(1))}{d(y,1)}\geq \sup_{1>y}1/y=1.$$ But for a contraction $f$ we must have $\sup_{x\ne y} \frac {d(f(y),f(x)}{d(y,x)}<1.$
(III). For $1\leq x<y$ we do have $d(f(x),f(y))<d(x,y),$ and $f$ does have a unique fixed point: $f(1)=1.$ For a function $f$ on a non-empty complete metric space to have a unique fixed point, it is sufficient that $f$ be a contraction, but it is not always necessary.