TITLE: Showing a contraction without a fixed point
QUESTION [6 upvotes]: Suppose $f: [1, \infty) \to [1, \infty]$ defined by $f(x) = x + \frac{1}{x}$ for all $x \geq 1$. I want to prove that:
\begin{equation}
|f(x)-f(y)| < |x-y|
\end{equation}
except when $x=y$, but $f$ does not have a fixed point.
By the Banach fixed point theorem we know that if a function $f: X \to X$ is a contraction of a complete metric space, then $f$ has a unique fixed point $p$ and the sequence of $(f, f \circ f, f\circ f\circ f, ...)$ that is the sequence of $f$ composed with itself $n$ times at index $n$ converges to p for all $x$.
But $[1, \infty]$ is not a complete metric space. So it seems like a good idea to proceed via contradiction? Where can I go from here
All help is greatly appreciated

REPLY [2 votes]: I want to extend kahen's comment.
It is in fact related to the contraction definition:
a map such that $$d(f(x), f(y)) < d(x,y)$$ is called contractive, whereas one such that there exists some $0<k<1$ such that $$d(f(x), f(y)) \leq kd(x,y)$$ $\forall x, y$ is called a contraction (which is required in Banach's theorem). 
[Note that contraction $\implies$ contractive but that the opposite is wrong]
The given function (where $d$ is given by $d(x, y) = |x-y|$) is a good counterexample showing that one can not relax the contraction condition in Banach's theorem to contractive. 
[Note that there is, however, a theorem by M. EDELSTEIN giving a sufficent condition on a contractive map to admit a fixed point]