This exercise is in the context of Lie Groups.
Exercise: Show that the collection $\{xU\}$, over all neighborhoods of $U$ of $e$, is a base of neighborhoods for $x$(similarly for $\{Ux\}$)
I thought of solving this question in the folloing way:
Consider:
$L:G\times U\to G$
such that $L(x,u)=xu$ such that $u\in U$
Fixing $x$, L is a diffeomorphism and $L(x,e)=x$ once $e\in U$, so $x\in Ux$, as well as the other points, forming a neighborhood.
Question
How do I prove the other points $L(x,u)$ will be in $\{xU\}$? It seems that is the definiton of $xU$.
Let's consider $L: G\times G\to G$ as the left translation, then $L$ is a diffeomorphism. L is obviously invective and subjective. But is it open? Does it have a continuously differentiable inverse? Bu Lie Group definition the operations $L_g(x)=gx$ and $L_{g^{-1}}(x)=g^-1 x$ are $C^{\infty}$. Noting that $L_{g^{-1}}\circ L_g(x)=g^{-1}gx=x$. The function has a $C^{\infty}$ inverse, hence it is open. Then $L_x(U)$ is an open set containing $x$ once $L_x(e)=x$, so xU is a neighbourhood of x. The same holds for every open set hence $\{xU\}$ constitutes the collection of all neighbourhoods. In a similar way, this applies to the right translation.