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This question already has an answer here:

Let $(X,d)$ be compact. Show: for a map $f$ that when $\forall x, y \in X$ with $x\neq y$

$d(f(x),f(y))<d(x,y)$ is fulfilled.

Then $f$ has a unique fixed point.

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marked as duplicate by user99914, John B, cansomeonehelpmeout, samerivertwice, José Carlos Santos real-analysis May 22 '18 at 17:37

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Assume there are two fixed points, let $x$ and $y$.

As $f(x)=x$ and $f(y)=y$, then $$d(f(x),f(y))=d(x,y)<d(x,y),$$ a contradiction.

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  • $\begingroup$ I don't see how compactness matters. $\endgroup$ – Yves Daoust May 22 '18 at 8:28
  • $\begingroup$ The statement $f$ has a unique fixed point means $f$ has a fixed point and it does not have more than one fixed point. Where did you prove the existence of a fixed point? $\endgroup$ – Kavi Rama Murthy May 22 '18 at 8:58
  • $\begingroup$ @KaviRamaMurthy: the title led me to think that only uniqueness was to be established. $\endgroup$ – Yves Daoust May 22 '18 at 9:02
  • $\begingroup$ Even the title talks about both existence and uniqueness. $\endgroup$ – Kavi Rama Murthy May 22 '18 at 9:40
  • $\begingroup$ Extremely sorry. I didn't know the the title was changed. $\endgroup$ – Kavi Rama Murthy May 22 '18 at 11:39

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