# Proof of partition of unity

Let $A\subset\mathbb R^n$ and let $O$ be an open cover of $A$. Then we can find $C^\infty$ partition of unity for $A$. Now Spivak starts of by considering the case that $A$ is compact (see below).

Now I do see that $\psi_1(x)+\dots+\psi_n(x)>0$ for all $x\in A$, but I don't understand why this also holds for some open $U$ containing $A$. Because all we know is that the $\psi_i$ are positive on $D_i$, but they might still be zero on some point in $U_i$ that doesn’t lie in $D_i$. Even if they mean $U=\{U_1,\dots,D_n\}$, then $U$ can still be greater than $A$, so how about those points that lie in $U\setminus A$?

Since $\psi=\psi_1+\cdots+\psi_n$ is continuous, $U=\psi^{-1}\left((0,+\infty)\right)$ is open and contains $A$.