Exercise 6, § 3

Let $G$ be a topological group satisfying the following conditions: $(\mathrm{LR}_I)$ $G$ is separated and locally connected. $(\mathrm{LR}_{II})$ There exists a connected neighbourhood $U$ of the identity element $e$ of $G$ such that the complement of $e$ with respect to $U$ is not connected.

Under these conditions, prove that $G$ is a topological group locally isomorphic to $\mathbf{R}$. (We shall first prove that, for every connected neighbourhood $V$ of $e$ in $G$ contained in $U$, $V \cap C\{e\}$ is not connected, that $e$ is adherent to every connected component of $V \cap C\{e\}$, and that this set has exactly two connected components by reasoning as in Exercise 4. Then take the connected neighbourhood $V$ sufficiently small and define on $V$ a structure of a totally ordered set, such that $\mathscr{T}_0(V)$ is less fine than the topology induced on $V$ by that of $G$, and such that prop. 2 of V, p. 9 is applicable to the set $E$ of the elements $\geq e$ of $V$.)