Exercise 4, § 3

Let $G$ be a topological group satisfying the following conditions: (R_I) $G$ is connected. (R_{II}) The complement $G^*$ of the identity element $e$ of $G$ is not connected. Under these conditions, 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 $\mathbf{R}$. We shall establish successively the following properties:

a) Let $(U_i)_{1 \leq i \leq n}$ be a partition of $G^*$ formed by sets open in $G^*$ ($n \geq 2$). Now prove that each of the sets $U_i$ is open in $G$, that $e$ is adherent to each of them and that $G$ is separated. Deduce that the closures $\overline{U}_i = U_i \cup \{e\}$ in $G$ are connected (I, p. 115, exerc. 4).

b) Let $A$ be a connected component of $U_i$; now prove that for every index $j \neq i$, one has $A^{-1} \overline{U}_j = A^{-1}$ (observe that $A^{-1} \overline{U}_j$ is connected and contains $A^{-1}$, and that $A^{-1} \overline{U}_j \subset G^*$ for $j \neq i$).

c) Now prove that necessarily $n = 2$ (take for each index $i$ such that $1 \leq i \leq n$, a connected component $A_i$ of $U_i$; for $j \neq i$, one has, by b), $A_i^{-1} A_j \subset A_i^{-1}$, $A_j^{-1} A_i \subset A_i^{-1} \subset U_j^{-1}$, whence $A_j^{-1} \subset U_j$). Deduce that $G^*$ has exactly two connected components $A, B$, that $B = A^{-1}$ and $\overline{A} = C(A^{-1})$.

d) The relation $yx^{-1} \in \overline{A}$ is an order relation making $G$ a totally ordered group (show that $\overline{A}^2 = \overline{A}$ and that $x \overline{A} x^{-1} = \overline{A}$ whatever $x \in G$).

e) The topology $\mathscr{T}_0(G)$ is less fine than the given topology $\mathscr{T}$ on $G$ (observe that $A$ is open for $\mathscr{T}$).

Conclude with the aid of exerc. 2 above and of I, p. 82, prop. 4. (Cf. VI, p. 23, exerc. 12b).)

f) Now prove that if one supposes in addition that $G$ is locally compact, or locally connected, $G$ is isomorphic to $\mathbf{R}$.