Let $U$ be an open subset of $\mathbb{R}^n$, and suppose I have a vector field $F: U \to \mathbb{R}^n$. Does there exist necessary and sufficient conditions on $U$ such that there is an 'as nice as $F$' vector field $\bar{F}: \bar{U} \to \mathbb{R}^n$, where $\bar{U}$ is the closure of $U$? In particular, is there a condition on $U$ that holds for all such vector fields within a certain class, say smooth.
Clearly if $U$ is a punctured open ball, for example, problems arise such as the usual counterexample of $d\theta$ on $\mathbb{R}^2 \setminus \{0\}$, which suggests some condition in the spirit of $U = \textrm{int}(\bar{U})$, but this seems likely insufficient generally.