Let \(f: X \to X\) be continuous. Show that if \(X = [0, 1]\), there is a point \(x\) such that \(f(x) = x\). The point \(x\) is called a fixed point of \(f\). What happens if \(X\) equals \([0, 1)\) or \((0, 1)\)?
Solution
For \(X=[0,1]\), define \[ g(x)=f(x)-x. \] This is continuous. Since \(f(0)\in[0,1]\) and \(f(1)\in[0,1]\), \[ g(0)=f(0)\ge0,\qquad g(1)=f(1)-1\le0. \] By the intermediate value theorem, \(g(c)=0\) for some \(c\in[0,1]\), hence \(f(c)=c\).
The conclusion fails for \([0,1)\) and \((0,1)\). In either case the map \[ f(x)=\frac{x+1}{2} \] maps the space into itself and satisfies \(f(x)>x\) for every \(x<1\), so it has no fixed point.