Proposition 1

All non-empty open intervals of $\mathbf{R}$ are homeomorphic to $\mathbf{R}$.

Consider first a bounded open interval $I = ]a, b[$ ($a < b$). For each $x \in I$ put $f(x) = -\left( \frac{1}{x-a} + \frac{1}{x-b} \right)$. This function is continuous and strictly increasing on $I$, for we have seen that $\frac{1}{x-b}$ is strictly decreasing in $]-, b[$ and $\frac{1}{x-a}$ strictly decreasing in $]a, \to[$. It follows that $f$ is a homeomorphism of $I$ onto an interval $f(I)$ of $\mathbf{R}$ (§ 2, no. 6, Theorem 5). $f(I)$ is neither bounded above nor below; if, for example, we had $f(x) \leq c$ for all $x \in I$, it would follow that $1 - \frac{b-x}{x-a} \leq c(b-x)$, since $b-x > 0$; and this leads to a contradiction when $x$ is sufficiently near $b$ (by virtue of the continuity of the two sides of the inequality, which are rational functions, at the point $b$). Hence $f(I) = \mathbf{R}$, and therefore every bounded open interval is homeomorphic to $\mathbf{R}$. Let $g$ be the inverse of $f$: it maps every unbounded open interval of $\mathbf{R}$ onto an interval $J$ contained in $I$ and open in $I$. Since $I$ is open in $\mathbf{R}$, $J$ is also open in $\mathbf{R}$; since it is bounded, it is homeomorphic to $\mathbf{R}$; and thus we have proved that every unbounded open interval is homeomorphic to $\mathbf{R}$.