I wanted to ask the following: If I have a topological group $G$, I know I can create a base for a uniform space as follows: for each $U$ a neighborhood of e, we define $V_u= \{(x,y):x^{-1}y\in U\}$. We do that for every $U$ in a neighborhood base of $e$. Another notation is for $F\subset X$. So $y\in (V_u)^m[F]$ if there exist $f,x_1,x_2,...,x_{m-1},y$ s.t. $f^{-1}x_1,x_1^{-1}x_2,...,x_{m-1}^{-1}y\in V_u$.
I would like to know why $FU^m=(V_u)^m[F]$. I could only show $(\supseteq)$ but couldn't show the other direction. If you could help I would much appreciate it.
If $W$ is an entourage of a uniform space $(X,\mathscr{U})$, we define $W[x] = \{ y : (x,y) \in W\}$ for points $x\in X$, and $W[M] = \{ y : (\exists x\in M)((x,y)\in W)\} = \bigcup_{x\in M} W[x]$. Further, for entourages $W_1,W_2$, we have $W_2\circ W_1 = \{ (x,z) : (\exists y)((x,y) \in W_1 \land (y,z)\in W_2)\}$. Then it is easy to check that $$(W_2\circ W_1)[M] = W_2(W_1[M]).\tag{$\ast$}$$
Now we see that $V_U[F] = F\cdot U$:
\begin{align} V_U[F] &= \{ y : (\exists f\in F)((f,y) \in V_U\}\\ &= \{ y : (\exists f\in F)(f^{-1}y \in U)\}\\ &= \{ y : (\exists f\in F)(\exists u\in U)(y = f\cdot u)\}\\ &= F\cdot U. \end{align}
And the rest is iterated application of $(\ast)$:
$$(V_U)^{m+1}[F] = V_U[(V_U)^m[F]] = V_U[F\cdot U^m] = (F\cdot U^m)\cdot U = F\cdot U^{m+1}.$$