;\;}\|f-{\mathrm{id}}\|<{\varepsilon}\}\subset G$ centered at the identity homeomorphism ${\mathrm{id}}_{\mathbb R}$ will be denoted by $B({\varepsilon})$.

1\. We need to show that for every $n\in{\omega}$ the triple $(G_{n+2},G_{n+1},G_n)$ is balanced. This will follow as soon as we check that for every neighborhood $U\subset G_{n+1}$ of the identity homeomorphism ${\mathrm{id}}_{\mathbb R}$ and any neighborhood $W\subset G_{n+2}$ of ${\mathrm{id}}_{\mathbb R}$ the set $U\cdot\sqrt[G_n]{W}$ is a neighborhood of ${\mathrm{id}}_{\mathbb R}$ in $G_{n+2}$. Since the Whitney topology on the subgroup $G_{n+2}={\mathcal H}_+({\mathbb I}_{n+2})$ is generated by the sup-metric, the neighborhood $W\subset G_{n+2}$ contains the ${\varepsilon}$-ball $G_{n+2}\cap B({\varepsilon})$ for some positive constant ${\varepsilon}<1$. The constant ${\varepsilon}$ can be chosen so small that $G_{n+1}\cap B({\varepsilon})\subset U$.

Consider the closed subgroup $$H=\{h\in G_{n+2}{\, ;\;}{\mathrm{supp}}(h)\subset{\mathbb I}_{n+2}\setminus{\mathbb I}_{n}\}$$ of $G_{n+2}$ and observe that $W\cap H\subset \sqrt[G_n]{W}$. Now it suffices to check that $U\cdot (W\cap H)$ contains the ball $G_{n+2}\cap B({\varepsilon}/2)$. Take any homeomorphism $h\in G_{n+2}\cap B({\varepsilon}/2)$ and observe that $h$ maps the interval ${\mathbb I}_n=[-n,n]$ into the interval $[-n-{\varepsilon}/2,n+{\varepsilon}/2]$. So, we can consider the homeomorphism $g\in G_{n+1}$, which is equal to $h$ on the interval ${\mathbb I}_n$ and is linear on the intervals $[n,n+1]$ and $[-n-1,-n]$. It is clear that $\|g-{\mathrm{id}}\|\le\|h-{\mathrm{id}}\|<{\varepsilon}/2$ and $\|g^{-1}-{\mathrm{id}}\|=\|g-{\mathrm{id}}\|$. Let $f=h\circ g^{-1}\in G_{n+2}$. The equality $g|{\mathbb I}_n=h|{\mathbb 