I'm asked to prove the following statement:
Let $f$ be a contraction mapping on a complete metric space $M$ in the sense that $$ d(f(x), f(y)) \leq c d(x, y), \quad \forall x, y \in M $$ for some $c \in(0,1) .$ Let $x_{1} \in M,$ and define $x_{n+1}=f\left(x_{n}\right)$ for $n=1,2, \ldots$ Prove that $x_{n}$ is convergent and its limit is the unique solution to the equation $f(x)=x$
I think I should use the Cauchy sequence to prove it. I'm trying something like $\forall m,n>N, d(x_m,x_n) \leq cd(x_{m-1},x_{n-1}) \leq \dots \leq c^{m-1}d(x_1,x_{m-n})$, but I'm not sure if it is the right way.