How is $G_1 = \cup_{k=1}^{\infty} U^k$ a closed subgroup of $G^0$? for Lie group $G$

Question

Let $G$ be a Lie group, and $G^0$ the identity component. Let $U$ be an open subset of $G^0$ containing the identity such that $U = U^{-1}$. Let $$ G_1 = \cup_{k=1}^{\infty} U^k. $$ Clearly $G_1$ is open, but it is also supposed to be closed. I would appreciate an explanation of why this should be closed. Thank you.