I}_n$ implies $f|{\mathbb I}_n={\mathrm{id}}|{\mathbb I}_n$ and thus $f\in H$. It follows that $\|f-{\mathrm{id}}\|=\|h\circ g^{-1}-{\mathrm{id}}\|\le\|h\circ g^{-1}-g^{-1}\|+\|g^{-1}-{\mathrm{id}}\|<{\varepsilon}/2+{\varepsilon}/2={\varepsilon}$. Now we see that the elements $g\in G_{n+1}\cap B({\varepsilon}/2)\subset U$ and $f=g^{-1}\circ h\in H\cap B({\varepsilon})\subset \sqrt[G_n]{W}$ yield $h=g\circ f\subset U\cdot\sqrt[G_n]{W}$, which establishes the required inclusion $G_{n+2}\cap B({\varepsilon}/2)\subset U\cdot\sqrt[G_n]{W}$.

2\. Assuming that the tower $(G_n)_{n\in{\omega}}$ satisfies ${\mathrm{PTA}}$, we can find a neighborhood $U\subset G_1$ of ${\mathrm{id}}_{\mathbb R}$ such that for every neighborhood $V\subset G_2$ there is a neighborhood $W\subset G_2$ such that $WU\subset UV$.

Find ${\varepsilon}\in(0,1)$ such that $U\supset G_1\cap B({\varepsilon})$. Then for the neighborhood $V=G_2\cap B({\varepsilon})$ there is a neighborhood $W\subset G_2$ of ${\mathrm{id}}_{\mathbb R}$ with $WU\subset UV$. Find a positive constant $\delta<{\varepsilon}$ with $G_2\cap B(\delta)\subset W$. It follows that $(G_2\cap B(\delta))\cdot (G_1\cap B({\varepsilon}))\subset WU\subset UV\subset G_1\cdot (G_2\cap B({\varepsilon}))$ and after inversion, $(G_1\cap B({\varepsilon}))\cdot(G_2\cap B(\delta))\subset (G_2\cap B({\varepsilon}))\cdot G_1$. Take a homeomorphisms $f\in G_2\cap B(\delta)$ such that $f(1)=1-\delta/2$ and a homeomorphism $g\in G_1\cap B({\varepsilon})$ such that $g(1-\delta/2)=1-{\varepsilon}$. Then $g\circ f(1)=1-{\varepsilon}$ which is not possible as $g\circ f\in (G_2\cap B({\varepsilon}))\cdot G_1\subset\{h\in G_2:|h(1)-1|<{\varepsilon}\}$.

3\. Since the tower $(G_n)_{n\in{\omega}}$ is balanced, the topology of ${\mathrm g\mbox{-}\kern-2pt\varinjlim}G_n$ coincides with the topologies $\LA[\tau]$, $\RA[\tau]$, $\LR[\tau]$, and $\RL[\tau]$ according to Theorem \[t4.2\]. Since ${\mathrm g\mbox{-}\kern-2pt\varinjlim}G_n$ carries the strongest group topology inducing the 