Proposition 1

Let G be a topological group and A and B subgroups of G. Then $(\overline{A}, \overline{B}) = (\overline{A}, \overline{B}), \overline{D^iA} = \overline{D^iA}, \overline{C^iA} = \overline{C^iA}$.

Let $\phi$ be the continuous mapping $(x, y) \mapsto x^{-1}y^{-1}xy$ of $G \times G$ into G. Then $\phi(A \times B) \subset (A, B)$, hence $\phi(\overline{A} \times \overline{B}) \subset (\overline{A}, \overline{B})$ and hence

the opposite inclusion is obvious and therefore $(\overline{A}, \overline{B}) = (\overline{A}, \overline{B})$. Clearly $\overline{D^0A} = \overline{D^0A}$; assuming the equality $\overline{D^iA} = \overline{D^iA}$, it follows that

and hence $\overline{D^iA} = \overline{D^iA}$ for all $i$. The proof of the formula $\overline{C^iA} = \overline{C^iA}$ is analogous.