A strict contraction on a complete metric space has exactly one fixed point, found by iteration.Suggest a slogan
Let \((X,d)\) be a nonempty complete metric space, and let \(f\colon X\to X\) be a map for which there exists \(c\in[0,1)\) with \[ d(f(x),f(y)) \leq c\, d(x,y) \quad\text{for all } x, y\in X. \] Then \(f\) has a unique fixed point: there is exactly one \(x_0\in X\) with \(f(x_0) = x_0\).
Fix \(x\in X\) and set \(x_k\coloneqq f^k(x)\). By induction \(d(x_{k+1},x_k)\le c^kd(x_1,x_0)\), so for \(m>k\) the triangle inequality gives \(d(x_m,x_k)\le\frac{c^k}{1-c}d(x_1,x_0)\), and \((x_k)\) is Cauchy. By completeness \(x_k\to x^*\) for some \(x^*\in X\); since \(f\) is continuous, \(f(x^*)=\lim_k f(x_k)=\lim_kx_{k+1}=x^*\). If \(f(y)=y\) as well, then \(d(x^*,y)=d(f(x^*),f(y))\le c\,d(x^*,y)\) with \(c<1\), so \(d(x^*,y)=0\) and \(y=x^*\).