Problem with finding a suitable partition of unity Let $\{U_s\}_{s\in S}$ be a family of open subsets  of $\mathbb R$ with union $W$  with the following property: for each $x\in U_s$ we have $x+1, x-1\in U_s$. 
Does there exist a sequence $\phi_1,\phi_2, \ldots$ of  $1$-periodic, smooth, compactly supported nonnegative functions such that:


*

*for each $n\in \mathbb N$ there is a $s(n)\in S$ such that $supp \phi_n \subset  U_{s(n)}$

*$\sum_{n=1}^\infty \phi_n(x)=1$ for $x\in \mathbb R$,

*for each compact $K\subset W$ there is an open set $G$ such that $G$ intersects only finite among sets $supp \phi_1, supp \phi_2,\ldots$.
 A: Since the closed interval $[0,1]$ is compact, finitely many of $U_s$'s, say $U_{s_1},\cdots, U_{s_m}$ , can cover it. Because of the assumption, these open sets cover whole $\mathbb{R}$ too. So they make an open cover for $\mathbb{R}$. 
Because of your assumption the image of elements of these open cover under the quotient map $\mathbb{R}\to \mathbb{T}$, $t\mapsto e^{2\pi it}$ is an open cover for the circle $\mathbb{T}$.
Consider a partition of unity subordinate to these smaller open cover for $\mathbb{T}$, say $\psi_1,\cdots, \psi_m$. The pullback of this partition of unity under the above quotient map is the answer. By pullback of $\psi_i$, I mean the function defined by $\phi_i(x):=\psi_i(e^{2\pi ix})$ for all $i=1,\cdots,m$.
Remember such a partition of unity has to be finite and the last part of the question automatically follows from the construction. 
Edit: The above answer is correct only if $W=\mathbb{R}$, and only in this case the smaller open cover is finite. When $W\neq \mathbb{R}$, you can check that the image of $W$ under the quotient map, say $\overline{W}$, is an open subset of $\mathbb{T}$ and instead of a finite cover for $\mathbb{T}$, you need to consider an infinite cover for $\overline{W}$ and find a subordinate partition of unity with respect to this infinite open cover, say $\{ \psi_i\}_{i=1}^\infty$ and then take the pullback of this partition of unity. 
