Let $G$ a topological group, $U \neq \emptyset$ open subset of $G$, $D$ dense in $G$. Prove $G=DU=UD$.
Attempt 1 Take $V=U \cap U^{-1}$. $V$ is open set. Let $x \in G$ then $xV$ is open and $D$ dense, so $xV \cap D \neq \emptyset$. Then exists $v \in V, d \in D$ such that $xv=d$ then $x=dv^{-1} \in DV^{-1} \subset DV \subset DU. $ But the problem is that $V$ empty because for all $y \in U$ is not necessarily $y \in U^{-1}$.
Attempt 2. If $e \in U$, $e$ is identity of group. Then exists $V$ a neighborhood of $e$ such that $V=V^{-1} \subset U$. Let $x \in G$ then $xV$ is a neighborhood of $x$ then $xV \cap D \neq \emptyset$, then $xv=d$ for some $v \in V, d \in D$ then $x=dv^{-1} \in DV^{-1} \subset DV \subset DU.$ But just se know that $U \neq \emptyset$ but I don't know how is possible $e \in U$.