Problem taken from my first semester analysis course homework
PROBLEM
Let $(X,d)$ be a compact metric space and $f:X\rightarrow X$ be a mapping such that $d(f(x),f(y)) < d(x,y)$ for all $x\neq y$. Show that there exists a unique $x_0\in X$ such that $f(x_0)=x_0$.
PROOF
For all $x,y\in X, x\neq y$ we have that $d(f(x),f(y)) < d(x,y)$. Thus $f:X \rightarrow X$ is continuous. Note that $g(x):=d(x,f(x))$ is also continuous. Recall that continuous functions functions on compact sets have a minimum on $X$. Thus let $d(x_0, f(x_0))=\inf\{d(x,f(x)):x\in X\}$. Assume by way of contradiction that $x_0\neq f(x_0)$. Then $d(f(x_0), f(f(x_0))) < d(x_0,f(x_0))$ which contradicts the fact that $g$ has a minimum at $x_0$. To see uniqueness, assume there exists $y\neq x_0$ where $f(y)=y$. But then $d(x_0, y)=d(f(x_0),f(y)) < d(x_0,y)$ which is a contradiction.
PROBLEM
Let $(X,d)$ be a compact metric space and $f:X\rightarrow X$ be a mapping such that $d(f(x),f(y)) < d(x,y)$ for all $x\neq y$. Show that there exists a unique $x_0\in X$ such that $f(x_0)=x_0$.
PROOF
For all $x,y\in X, x\neq y$ we have that $d(f(x),f(y)) < d(x,y)$. Thus $f:X \rightarrow X$ is continuous. Note that $g(x):=d(x,f(x))$ is also continuous. Recall that continuous functions functions on compact sets have a minimum on $X$. Thus let $d(x_0, f(x_0))=\inf\{d(x,f(x)):x\in X\}$. Assume by way of contradiction that $x_0\neq f(x_0)$. Then $d(f(x_0), f(f(x_0))) < d(x_0,f(x_0))$ which contradicts the fact that $g$ has a minimum at $x_0$. To see uniqueness, assume there exists $y\neq x_0$ where $f(y)=y$. But then $d(x_0, y)=d(f(x_0),f(y)) < d(x_0,y)$ which is a contradiction.
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