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Let (X,d) be a compact metric space and $ f \colon X \to X $ a uniformly continuous function so that $ d( f(x),(f(y)) < d(x,y) , \forall x,y \in X , x \neq y $ . Prove that there exists x' in X so that $ f(x') = x'.$ I would choose any x and f(x) in X and then look at $ d( f(x),f(f(x))$ and then $ d( f(f(x)),f(f(f(x)))$ and so on . Because of the assumtion this would lead to zero. Edit : found the answer this is a duplicate

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marked as duplicate by Robert Z, Saucy O'Path, Paul Frost, user10354138, Rebellos Nov 17 '18 at 13:58

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  • $\begingroup$ $f$ should be a function from $X$ to $X$. $\endgroup$ – Joppy Nov 17 '18 at 11:12

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