Exercise 4, § 3

Let $G$ be a topological group which satisfies the following conditions: (R_I) $G$ is connected. (R_{II}) The complement $G^*$ of the identity element $e$ of $G$ is not connected.

Then there exists a continuous bijective homomorphism of $G$ onto $\mathbf{R}$ (in other words, $G$ is algebraically isomorphic to $\mathbf{R}$ and its topology is finer than that of $\mathbf{R}$). The proof is in several steps:

a) Let $(U_i)_{1 \leq i \leq n}$ be a partition of $G^*$ into open sets in $G^*$ ($n \geq 2$). Show that each $U_i$ is open in $G$, that $e$ lies in each $\overline{U}_i$, and that $G$ is Hausdorff. Deduce that the closures $\overline{U}_i = U_i \cup \{e\}$ in $G$ are connected (Chapter I, § 11, Exercise 4).

b) Let $A$ be a connected component of $U_i$. Show that for each index $j \neq i$, we have $A^{-1} \overline{U}_j = A^{-1}$ (observe that $A^{-1} \overline{U}_j$ is connected and contains $A^{-1}$, and that $A^{-1} \overline{U}_j \subset G^*$ if $j \neq i$).

c) Show that $n$ must be equal to 2 [for each index $i = 1, 2, \ldots, n$ take a component $A_i$ of $U_i$; if $j \neq i$ we have, by b), $A_i^{-1} A_j \subset A_i^{-1}$, $A_j^{-1} A_i \subset A_j^{-1} \subset U_j^{-1}$, whence $A_i^{-1} \subset U_j$]. Deduce that $G^*$ has exactly two components $A$ and $B$, that $B = A^{-1}$ and $\overline{A} = C(A^{-1})$.

d) The relation $y x^{-1} \in \overline{A}$ is an order relation, which makes $G$ into a linearly ordered group (show that $\overline{A}^2 = \overline{A}$ and that $x \overline{A} x^{-1} = \overline{A}$ for all $x \in G$).

e) The topology $\mathscr{T}_0(G)$ is coarser than the given topology $\mathscr{T}$ on $G$ (observe that $A$ is open with respect to $\mathscr{T}$).

Complete the proof by using Exercise 2 above and Chapter I, § 11, no. 2, Proposition 4. [Cf. Chapter VI, § 1, Exercise 12 b).]

f) Show that if in addition we suppose $G$ to be either locally compact or locally connected, then $G$ is isomorphic to $\mathbf{R}$.