is a topological group. Since $\RL[G]$ is a quasitopological group, it suffices to check the continuity of multiplication at the neutral element. Given a basic neighborhood ${\operatornamewithlimits{\overleftrightarrow{\textstyle\prod}}}_{n\in{\omega}}W_n\in\RL[{\mathcal B}]_e$, we should find a neighborhood ${\operatornamewithlimits{\overleftrightarrow{\textstyle\prod}}}_{n\in{\omega}}V_n\in\RL[{\mathcal B}]_e$ such that $\big({\operatornamewithlimits{\overleftrightarrow{\textstyle\prod}}}_{n\in{\omega}}V_n\big)^2\subset {\operatornamewithlimits{\overleftrightarrow{\textstyle\prod}}}_{n\in{\omega}}W_n$.

For every $n\in{\omega}$, find a symmetric neighborhood $U_n\subset G_n$ of $e$ such that $U_n^2\subset W_n$ and let $Z_{n}=\sqrt[G_{n-2}]{U_{n}}$ be the maximal $G_{n-2}$-invariant subset of $U_{n}$ (here we assume that $G_k=\{e\}$ for $k<0$). Let $\widetilde U_0=W_0$ and by induction for every $n\in{\mathbb N}$ choose a symmetric neighborhood $\widetilde U_n\subset U_n$ of $e$ such that $\widetilde U_n^3\subset Z_n\widetilde U_{n-1}Z_n$. The choice of the neighborhood $\widetilde U_n$ is possible because the set $Z_n\widetilde U_{n-1}Z_n$ is a neighborhood of $e$ in the group $G_n$ by the bi-balanced property of the triple $(G_n,G_{n-1},G_{n-2})$. Finally, for every $n\in{\omega}$ let $V_n=G_n\cap \widetilde U_{n+1}$.

We claim that ${\operatornamewithlimits{\overleftrightarrow{\textstyle\prod}}}_{n\in{\omega}}V_n$ is the required neighborhood with $\big({\operatornamewithlimits{\overleftrightarrow{\textstyle\prod}}}_{n\in{\omega}}V_n\big)^2\subset {\operatornamewithlimits{\overleftrightarrow{\textstyle\prod}}}_{n\in{\omega}}W_n$. This inclusion will follow as soon as we check that $$\label{SIN2b}
\big({\operatornamewithlimits{\overleftrightarrow{\textstyle\prod}}}_{n<m}V_n\big)^2\subset {\operatornamewithlimits{\overleftrightarrow{\textstyle\prod}}}_{n\le m}W_n$$ for all $m\in{\omega}$.

For $m=0$ this inclusion is trivial. Assume that the inclusion (\[SIN2b\]) has been proved for some $m=p\in{\omega}$. We shall prove it for $m=p+1$. For every non-negative $k<m$ consider the subset $$\Pi_{k}=\big({\operatornamewithlimits{\overleftarrow{\textstyle\prod}}}_{k\le n<m}V_nZ_{n+1}\big)\cdot\big({\operatornamewithlimits{\overleftrightarrow{\textstyle\prod}}}_{n<k}V_n\big)\cdot \widetilde U_{k}\cdot \big({\operatornamewithlimits{\overleftrightarrow{\textstyle\prod}}}_{n< k}V_n\big)\cdot \big({\operatornamewithlimits{\overrightarrow{\textstyle\prod}}}_{k\le 