The bounded sequence $(x_n)$ lies in a compact set because $X$ is a <proper metric space>. After taking a subsequence, $x_n\to x$. Since $d(x_n,\phi_nx_n)\to0$, both $x_n$ and $\phi_nx_n$ eventually lie in one compact neighborhood $L$ of $x$. A <properly discontinuous group action> has only finitely many $g\in\Gamma$ with $gL\cap L\ne\varnothing$, so some conjugate $\psi$ occurs as $\phi_n$ along an infinite subsequence. Continuity then gives
$$
d(x,\psi x)=\lim_n d(x_n,\psi x_n)=0.
$$
Thus $\psi$ fixes $x$. Since $\psi=\gamma\phi\gamma^{-1}$ for some $\gamma$, the original element $\phi$ fixes $\gamma^{-1}x$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 133 / 4 / c / Solution

{wiki=Properly_discontinuous_action}

An action is properly discontinuous when every compact set meets only finitely many of its translates. For an isometric action on a proper metric space, this gives the finiteness needed to extract a constant group element from a sequence with bounded source and target.


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 Incoming links: Properly discontinuous group action