Let (X, d) be a complete metric space and 0 < < 1. Moreover, let f : X X such that for all x, y X it holds that d(f(x), f(y)) d(x, y). We will show that there exists a unique x X such that f(x) = x using the following steps. This statement is called Banach's xed point theorem and has numerous applications, for example to show existence of solutions to a relatively general class of ordinary dierential equations. (a) (2 points) Let x X. Dene a sequence by x1 = x, xn 1 = f(xn). Show for n N it holds that d(xn 1, xn) n1d(x1, x2). (b) (2 points) Use a) to show that (xn)nN is a Cauchy sequence. Hint: Use the triangle inequality and the fact that sn = Pn k=0 k is convergent and therefore Cauchy. (c) (2 points) Show that if a sequence (yn)nN in X converges to y X then f(yn) f(y). (d) (2 points) Conclude that there exists x X such that f(x) = x. (e) (2 points) Prove that the so-called xed point from d) is unique