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We have a complete metric space $(X,d)$ with a surjective function $f:X \to X$. There also exists a $c>1$ in order that for all $x,y \in X$
$d(f(x),f(y)) ≥ cd(x,y)$
I have proved that $f$ is injective and with help of the inverse function that $f$ has a unique fixed point with the fixed point theorem. I now have to show that it is necessary for $f$ to be surjective but I don't know how I can show this, any suggestions?

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  • $\begingroup$ I suspect that $X$ is compact.. $\endgroup$ – Red shoes Jun 17 '17 at 10:48
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You need to find a counterexample. Here is one.

Let $X = [0,\infty)$ with the absolute-value metric and $f(x) = 2x+1$.

Exercise: Show $f$ is not surjective.

Let's see what the inequality looks like. . . .

$d(f(x),f(y)) = d(2x+1,2y+1) = |(2x+1)-(2y+1)|=|2x-2y|=2|x-y| = 2d(x,y)$.

So the inequality holds for $c=2$.

But the function has no fixed point! (Exercise).

Therefore we cannot omit the hypothesis that $f$ is surjective.

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Take $f\colon\mathbb{Z}\longrightarrow\mathbb Z$ defined by $f(m)=3m+1$. The space $\mathbb Z$ is complete (with respect to the usual distance) and if $m,n\in\mathbb Z$, then $d\bigl(f(m),f(n)\bigr)=3d(m,n)$. But $f$ has no fixed point.

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Consider $X=[1,+\infty)$ with the Euclidean metric, and the function $f(x) = 2x$.

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