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Let $(X,d)$ be a compact metric space and let $f : X → X$ be a function with the property that $d( f (x), f (y)) < d(x, y)$ whenever $x \neq y$. Show that $f$ has a fixed point, that is, there exists $x_0$ such that $f (x_0) = x_0$. (Hint: consider the function $g(x) = d(x, f (x))$ and argue that it attains its minimum and the minimum is 0)

Thank you all.

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    When you followed the given hint, at which point did you have a problem? – 2017-01-18
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    $X$ compact, then $X$ is bounded. So I know $x\in X, f(x)\in X, \exists k$ s.t $d(x,f(x))=k$ is min. But I do not know how to say the min is 0 based on $d(f(x),f(y)) – 2017-01-18
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    Okay. So suppose you had $k > 0$. Then in particular $x \neq f(x)$. Which of the given properties might you want to apply in this situation? – 2017-01-18
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    $d(f(x),f(f(x))) – 2017-01-18
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    Hole in one. Bingo. Got it. – 2017-01-18
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    Wait, I did not get it. Why? – 2017-01-18
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    That's a contradiction to $$d(x,f(x)) = \min \{ d(y,f(y)) : y \in X\}.$$ – 2017-01-18
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    You've found a smaller value than the minimum. PS: It is not that $X$ is bounded that guarantees the existence of a minimum as you seem to imply. – 2017-01-18
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    @AloizioMacedo Then how can I guarantee the existence of min? – 2017-01-18
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    That follows from the compactness of $X$ (and the continuity of $g$). – 2017-01-18
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    @DanielFischer Like $g: X\rightarrow K$ is continuous and $X$ is compact, then $K$ is compact, so we have the min? – 2017-01-18
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    Well, if $g(X) = K$, then $K$ is compact. If $g(X) \subsetneqq K$, then $K$ could be anything. – 2017-01-18
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    Ok, I think we know $g(X)= K$. One more question, how can I know $g$ is continuous? We do not know if $f$ is continuous. – 2017-01-18
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    We have $d(f(x),f(y)) \leqslant d(x,y)$ for all $x,y\in X$. That implies that $f$ is continuous. – 2017-01-18
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    @DanielFischer Hey Daniel. I still feel confused about how to prove $g$ is continuous. Even if now I know $f$ is continuous, how can I get that $g$ is continuous? – 2017-01-23
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    One part of it is that $d \colon X\times X \to \mathbb{R}$ is continuous, the other part is that $x \mapsto (x,f(x))$ is continuous. But if you don't know about the product topology yet, note that $$\lvert g(y) - g(x)\rvert = \lvert d(y,f(y)) - d(x,f(x))\rvert \leqslant d(y,x) + d(f(y),f(x)),$$ and that is $\leqslant 2 d(y,x)$. – 2017-01-23

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You can take $x_{n+1} = f(x_n)$. Then since $X $ is compact there exist a convergent subsequence $x_{n_l} = y_l$ whose limit $x_0$ belongs to $X$. Now since $x_{n_l} = y_l$ converges to $x_0$ therefore $d(f(x),f(y))< d(x,y) $ $ \implies f(y_l) \rightarrow x_0$. hence $f(x_0) = x_0$