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Let $(X,d)$ be a metric space. Let $T: X\to X$ be a contraction map. Then there is a $R>0$ such that $T(B(x, R) )$ is subset of $B(x,R)$.

Is it always true? My geometric intuition says the answer is yes. However I cannot proceed anymore. Thanks in advance. Any help would be appreciated.

Robert Z
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1 Answers1

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Do you mean that there is $R>0$ such that for all $x\in X$, $T(B(x, R) )\subseteq B(x,R)$?

Then the property does not hold. Take $X=(0,+\infty)$ with $d(x,y)=|x-y|$ and $T=x/2$. Then $T$ is a contraction and if $x>R$ $$T((x-R,x+R)=((x-R)/2,(x+R)/2)\not \subseteq (x-R,x+R).$$

Robert Z
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