Past exam of the metric spaces and complex analysis course of the University of Oxford / 2023 / 2 / c

By the permitted sequential compactness of the product metric space $X\times X$, the sequence of pairs $({\varphi }^{(k)}x,{\varphi }^{(k)}y)$ has a convergent subsequence, and hence a Cauchy subsequence. Choose two of its indices $r<s$ sufficiently late thatThe distance-increasing assumption iterates: $d({\varphi }^{(r)}u,{\varphi }^{(r)}v)\ge d(u,v)$. With $n=s-r\ge 1$, use $u=x,v={\varphi }^{(n)}x$, and then $u=y,v={\varphi }^{(n)}y$, to obtainBoth summands are nonnegative, soThe use of paired iterates ensures that the same positive return time works for both points.

Choose $n\ge 1$ as in part (i). Since distances between iterates are nondecreasing, the triangle inequality givesLetting $ϵ\downarrow 0$ and combining with the given reverse bound proves that $\varphi$ is an isometry:For surjectivity, fix $x\in X$ and use part (i) with $y=x$ and $ϵ=1/k$ to choose $n_k\ge 1$ satisfying $d(x,{\varphi }^{(n_k)}x)<1/k$. Put $z_k={\varphi }^{(n_k-1)}x$, interpreting ${\varphi }^{(0)}$ as the identity. By sequential compactness, a convergent subsequence satisfies $z_{k_j}\to z\in X$. The assumed continuity yieldsThus every $x$ has a preimage, and $\boxed{\varphi \text{ is surjective}}$. If $X$ is empty, both conclusions hold vacuously. This proves the distance-increasing self-map theorem for sequentially compact metric spaces.