original topology on each group $G_n$, we conclude that the Whitney topology is weaker that the topology $\LA[\tau]$. In order to show that these two topologies coincide, it suffices to check that each basic neighborhood ${\operatornamewithlimits{\overrightarrow{\textstyle\prod}}}_{n\in{\omega}}U_n$ of $e$ in the topology $\LA[\tau]$ is a neighborhood of ${\mathrm{id}}_{\mathbb R}$ in the Whitney topology. Here for every $n\in{\omega}$, $U_n$ is an open symmetric neighborhood of ${\mathrm{id}}_{\mathbb R}$ in the group $G_n={\mathcal H}_+({\mathbb I}_n)$. Since the Whitney topology on the subgroup $G_n$ is generated by the sup-metric, we can find a positive constant ${\varepsilon}_n<1/2$ such that $U_n\supset G_n\cap B({\varepsilon}_n)$.

Choose a continuous function ${\varepsilon}:{\mathbb R}\to(0,\tfrac12)$ such that $$\label{E1}
\sup\{{\varepsilon}(x){\, ;\;}x\in{\mathbb I}_{n}\setminus {\mathbb I}_{n-4}\}<{\varepsilon}_n/2\mbox{ \ for all $n\in{\omega}$}.$$ Here we assume that ${\mathbb I}_k=\emptyset$ for all negative $k$.

The function ${\varepsilon}$ determines the neighborhood $B({\varepsilon})=\{h\in G{\, ;\;}|h-{\mathrm{id}}_{\mathbb R}|<{\varepsilon}\}$ of the identity map ${\mathrm{id}}_{\mathbb R}$ in the Whitney topology.

We claim that $B({\varepsilon})\subset{\operatornamewithlimits{\overrightarrow{\textstyle\prod}}}_{n\in{\omega}}U_n$. Fix any homeomorphism $h\in B({\varepsilon})$ and for every $n\in{\mathbb N}$ consider the homeomorphism $h_n\in G_{n}$ such that $h_n|{\mathbb I}_{n-1}=h|{\mathbb I}_{n-1}$ and $h_n$ is linear on the intervals $[n,n+1]$ and $[-n-1,-n]$. For $n\le 0$ we put $h_n={\mathrm{id}}_{\mathbb R}$. It is clear that $h_{m}=h$ for some $m\in{\mathbb N}$.

For every $n\in{\omega}$ consider the homeomorphism 