Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 4 / 12F (Analysis II)

A complete metric space is one in which every Cauchy sequence converges. On $I=(0,1]$ defineThe map $x\mapsto 1/x$ is an isometry from $(I,d)$ onto the closed subset $[1,\infty )$ of $\mathbb{R}$, so $d$ is a metric and is complete. That map and its inverse are continuous in the ordinary Euclidean metrics, so it is also a homeomorphism. Therefore $d$ induces exactly the usual subspace topology on $I$.

If $Y$ is closed and $(y_n)$ is Cauchy in $Y$, completeness of $X$ gives $y_n\to y\in X$, and closedness gives $y\in Y$. Conversely, if $Y$ is complete and $y$ lies in its closure, choose $y_n\in Y$ with $d(y_n,y)<1/n$. This sequence is Cauchy, so it converges in $Y$; uniqueness of limits makes its limit $y$. Thus $Y$ is complete exactly when it is closed in $X$.

Choose $x\in X$ and iterate $x_{n+1}=f(x_n)$. The contraction mapping estimatemakes $(x_n)$ Cauchy by a geometric series. Its limit $x_0$ satisfies $f(x_0)=x_0$ because $f$ is Lipschitz. Two fixed points would satisfy $d(x_0,y_0)\le \lambda d(x_0,y_0)$, so they coincide. This proves the Banach fixed-point theorem.

Now suppose the final assertion failed. There would be $m_k,n_k\to \infty$ with $f(a_{m_k})=a_{n_k}$. Since $a_n\to a$ and $f$ is continuous, the two sides tend respectively to $f(a)$ and $a$, so $f(a)=a$. Uniqueness would give $a=x_0$, a contradiction. Hence the required $\ell$ and $m$ exist.