Exercise 3, Appendix 0

Let $G$ be a Hausdorff topological group which has a neighbourhood $V_0$ of the identity element $e$ on which $x \to x^{-1}$ is uniformly continuous (as a mapping $G_d \to G_d$: cf. Chapter III, § 3, Exercises 7 and 8). Let $(x_i)_{i\in I}$ be a multipliable family of points of $G$; show that, for each 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 slice of $I$ which does not meet $J_0$, we have $p_k \in U$. [Show first that there exists a finite subset $H_0$ of $I$ and a finite number of points $a_1, a_2, \ldots, a_q$ of $G$ such that, for every finite subset $L$ of $I$ contained in a slice of $I$ which does not meet $H_0$, we have $p_L \in a_k V'_0 a_k^{-1}$ for at least one index $k$, where $V'_0$ is a neighbourhood of $e$ such that $V'_0 V'_0 \subset V_0$. Consequently we may restrict ourselves to considering finite subsets contained in one of the open intervals whose end-points are two consecutive indices of $H_0$; use the argument of Exercise 2 a) and the uniform continuity of $x^{-1}$ on each of the neighbourhoods $a_k V_0 a_k^{-1}$.]

Deduce that under the same conditions we have $\lim x_i = e$ with respect to the filter of complements of finite subsets of $I$ (if $I$ is infinite).