Isn't $[1,\infty)$ complete?

Question

Let $S=[1,\infty)$. For example, it can be shown that $f: S\to S$ defined by $f(x) = x+1/x$ is a contraction mapping on $S$, yet it has no fixed points on $S$. Should this mean that $S$ is not complete (as per the Banach contraction principle)? But if it is not complete, then I can't think of a Cauchy sequence in $S$ which would not converge in $S$. Can you?

Answer 1

In the Banach fixed point theorem, one requires the existence of a constant $c < 1$ such that $d(f(x), f(y)) \leqslant c\cdot d(x,y)$ for all $x,y$. This condition is not satisfied here, for $x\neq y$ we have

$$\frac{f(y) - f(x)}{y-x} = 1 - \frac{1}{xy},$$

thus for every $c < 1$ we can find $x,y$ with $\lvert f(x) - f(y)\rvert > c\cdot \lvert x-y\rvert$.

The weaker condition that $d(f(x),f(y)) < d(x,y)$ for all $x\neq y$ is sufficient for the existence of a (unique) fixed point if the premise that $X$ be a complete metric space is strengthened to the premise that $X$ be compact.