Exercise 3, Appendix 2

Let $G$ be a separated topological group such that there exists a neighbourhood $V_0$ of the identity element $e$, in which $x \mapsto x^{-1}$ is uniformly continuous (as a mapping from $G_d$ into $G_d$; cf. III, p. 25, prop. 9). Let $(x_i)_{i \in I}$ be a multipliable family of points of $G$; now prove that, for every neighbourhood $U$ of $e$, there exists a finite subset $J_0$ of $I$ such that, for every finite subset $K$ of $I$ contained in a segment of $I$ which does not meet $J_0$, one has $p_K \in U$. (First prove that there exists a finite subset $H_0$ of $I$ and a finite number of points of $G$, $a_1, a_2, \ldots, a_q$ such that, for every finite subset $L$ of $I$ contained in a segment not meeting $H_0$, one has $p_L \in a_k V'_0 a_k^{-1}$ for at least one index $k$, $V'_0$ being a neighbourhood of $e$ such that $V'_0 V'_0 \subset V_0$. Then restrict oneself to considering the finite subsets contained in one of the open intervals whose endpoints are two consecutive indices of $H_0$; use the reasoning of Exercise 2 a), and the uniform continuity of $x^{-1}$ in each of the neighbourhoods $a_k V_0 a_k^{-1}$).

Deduce that, under the same conditions, one has $\lim x_I = e$ following the filter of the complements of the finite subsets of $I$ (if $I$ is infinite).