Banach Fixed Point Theorem problem contradiction I have a following problem. 

Let $X = R$, $d(x,y) = |x-y|$, $T(x) = \sqrt{x^2 + 1}$ but $T$ does not have a fixed point. Does this contradict Banach's Fixed Point Theorem? 

I know that if $X$ was compact space then it would imply the existence and uniqueness of a fixed point, but not sure what to do in this case ($R$ is not compact). I would appreciate any help. 
 A: To solve such a problem the first and foremost thing is to recall the result in question. 
So, what does the Banach Fixed Point Theorem say: 

Let $(X, d)$ be a non-empty complete metric space with a contraction mapping $T : X \to X$. Then $T$ admits a unique fixed-point $x^\ast$ in $X$.

Now $\mathbb{R}$ with $d(x,y)= |x-y|$ is a complete metric space.  And, your $T$ is a map from $\mathbb{R}$ to itself. 
Yet, is it a contraction? What even is a contraction? Let us check the same Wikipedia page to find: 

A map $T : X \to X$ is called a contraction mapping on $X$ if there exists $q \in [0, 1)$ such that $d(T(x),T(y)) \le q d(x,y)$ for all $x, y$ in $X$. 

Is this true for your map $T$? (It can be important to pay attention to the fact that just $d(T(x),T(y)) <  d(x,y)$ for all $x, y$ in $X$ is not sufficient but really the slightly stronger condition is needed.)
(Note: the answer ought to be "no," because 'clearly' there  should not be an example that contradicts the well-known theorem, and your map has no fix point and this condition is the last we would need to check to get one.)
A: No, $T(x)$ doesn't have the contraction property, consider $x_n=n$ and $y=0$, then $|x_n-0|=n$, but
$$|T(x_n)-T(0)|=|\sqrt{n^2+1}-1|={n^2+1-1\over \sqrt{n^2+1}+1}$$
$$={n^2\over \sqrt{n^2+1}+1}$$
With $\epsilon>0$ arbitrary and fixed we see for $n>\!>0$ we have
$$n(1+\epsilon)^{-1}={n^2\over\sqrt{(1+\epsilon)^2n^2}}< |T(x_n)-T(0)|$$
But then by taking $\epsilon$ sufficiently small, we see that there can be no $q\in [0,1)$ such that
$$|T(x)-T(y)|<q|x-y|$$
in fact we see that clearly $q\ne 0$, so if a candidate $0<q<1$ is given then $\epsilon =\min\{{1\over 2},q^{-1}-1\}$ works.
for all $x,y\in\Bbb R$, because for some $n$ we have $x=x_n, y=0$ is a counterexample, hence $T$ is not a contraction.
