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Let (X,p) be a compact metric space.Suppose that f X->X is a function such that, for all $x_1$,$x_2$ $\in$X, if $x_1\neq x_2$ then p(f($x_1$),f($x_2$))<$p($x1$,$x2$)$. Prove that there exists a unique point y which in X such that f(y)=y. (there is a hint below the question:Consider the function g:X->R defined by setting g(x)=p(x,f(x)).)

this is my thought: I know g(X) should be a closed and bounded set in the real numbers, since X is compact. so m<=g(X)<=M, if m=0, it's obvious that there exists a x in X such that f(x)=x, but if m>0, I want to prove if m>0, then X cannot be a compact set, I am trying to find an open covering which there are no finite open sets which can cover X. If I can find such an open covering, then I think I will finish the question, but I don't know how to find it. Can someone help me to find it. Or, can someone have other methods to solve this question? I am struggling with it, and have no idea how to continue? Hope someone can help me solve this question!

Shootforthemoon
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python3
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1 Answers1

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As you said, if $m = 0$ we're done, otherwise, suppose $m > 0$, then because $X$ is compact there is some $x$ such that $g(x) = m$, now if we let $y = f(x)$ we have that $$g(y) = g(f(x)) = \rho(f(x),f(f(x))) < \rho(x, f(x)) = m$$

Wich is a contradiction, so $m= 0$ and we're done.

aram
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