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Let $(X, d)$ be a compact metric space. Let $f : X \to X$ be such that $d(f (x), f (y)) < d(x, y)$ for all $x, y \in X$ with $x \neq y$. To show that $f$ has a fixed point, that is, there exists $x_0 \in X$ such that $f (x_0) = x_o$.

Finding it difficult to prove the existence of a fixed point.

Is the fixed point unique?

User8976
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1 Answers1

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Uniqueness: if $x,y$ are distinct fixed points, then $f(x)=x$ and $f(y)=y$. So $d(x,y)=d(f(x),f(y))$, which contradicts the hypothesis.

Existence: the map $x\mapsto d(x,f(x))$ is continuous (why?), so there exists an $x_0$ which achieves the minimum.
Now let $\epsilon:=d(x_0,f(x_0))$. If $\epsilon>0$ we have $x_0\neq f(x_0)$, so $d(f(x_0),f^2(x_0))<\epsilon$ by hypothesis, which contradicts the minimality of $\epsilon$. So $d(x_0,f(x_0))=0$, i.e. $f(x_0)=x_0$.

Mizar
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