smooth functions and partition unity

Take $\displaystyle U \subset \mathbb{R}^{n}$ where $\displaystyle U$ is bounded. Assume we have an open covering $\displaystyle U \subset \cup_{i=0}^{N}V_{i}$ where $\displaystyle V_{i} \subset U$ for each $\displaystyle i$. For each $\displaystyle V_{i}$ we have the mappings $\displaystyle v_{i} \in C^{\infty}(\bar{V_{i}})$. If we let $\displaystyle \{ \zeta_{i} \}_{i}^{N}$ be a smooth partition of unity subordinate to the open sets $\displaystyle \{V_{i}\}_{i=0}^{N}$ in $\displaystyle U$ and we define $\displaystyle v := \sum_{i=0}^{N}\zeta_{i}v_{i}$ then $\displaystyle v \in C^{\infty}(\bar{U})$.

I have two questions:

Is it not required that each $\displaystyle \zeta_{i}v_{i}$ have $\displaystyle \bar{U}$ as a domain to define $\displaystyle v := \sum_{i=0}^{N}\zeta_{i}v_{i}$?

Secondly. how do we show that $\displaystyle v \in C^{\infty}(\bar{U})$, why is this true?

Thanks!