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.
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 $f(x_0) = x_0$