Partition of Unity involved in proving that $C^{\infty}(U,V)$ is dense in $C_S^r(U,V)$, differential topology I have been reading Hirsch's Differential Topology , and here is giving a proof that if $U\subset \mathbb{R}^m$ and $V\subset\mathbb{R}^n$ are open sets then $C^{\infty}(U,V)$ is dense in $C^r_S(U,V)$. Now for this in the proof he claims that if we have a locally finite family of compact sets $\{K_i\}_{i\in I}$ that covers $U$ then there exists a $C^{\infty}$ partition of unity on $U$ such that $supp \lambda_i$ is compact, $K_i\subset supp \lambda_i$ and if $K_i\cap K_j=\emptyset \Rightarrow supp\lambda_i \cap supp\lambda_j =\emptyset$ . Now I am not being able to see why this is true , can anyone give me some help ?
I have tried to construct one explicitly but I had no success. My idea was to cover the compact set by finite number of coordinate charts and find for each $\alpha$ an open set $U_{\alpha}$ such that $K_{\alpha}\subset U_{\alpha}$ such that the $U_{\alpha}$ are locally finite and disjoint if they're associated compact sets are disjoint. However I don't know how to justify why this would be true. If I have these two properties I think I am able to do the rest because we just need to cover $K$ with a finite number of open balls and then it's easy to construct the partition of Unity explicitly.
New edit: To try and deal with the properties I was talking about using the fact that the $K_{\alpha}$ are locally finite , for each $x\in K_{\alpha}$ I can choose $V_{x}$ such that this neighborhood intersects a finite number of the compact sets. Now consider the new neighborhood where $V_x':=V-(\cup_{i=1,i\neq \alpha}^n K_i)$, where  the $K_i's$ are the compacts that it intersects and that are disjoint with $K_{\alpha}$. Now we can cover $K$ by a finite number of these $V_{x}'s$ . Although I don't think this works  if we want the open sets $V_x'$ locally finite.
 A: Here's one way, requiring only the standard existence theorem for partitions of unity (for any open cover there is a locally finite POU subordinate to that cover). The useful observation is that there is a locally finite open cover $\{\mathcal{O}_i\}_{i\in I}$ with $K_i\subseteq\mathcal{O}_i$.
Let the neighbors of $K_i$ denote the $K_j$ for which $K_i\cap K_j\neq\emptyset$. Since they are compact, each $K_i$ has only finitely many neighbors.
Define a covering
$$
\mathcal{O}_i=U\setminus\bigcup_{\substack{j\in I\\K_i\cap K_j=\emptyset}}K_j
$$
Note that $K_i\subseteq\mathcal{O}_i$, and $\mathcal{O}_i$ is open, since it is the compliment of a locally finite union of closed sets. Furthermore, the set $\{\mathcal{O}_i\}_{i\in I}$ is locally finite. To see this, note that an open set $W\subseteq U$ intersects $\mathcal{O}_i$ only if it intersects a neighbor of $K_i$.
Now, consider the open cover $\{\mathcal{O}_i,U\setminus K_i\}$, and choose a partition of unity $\{\psi_i,\varphi_i\}$ subbordinate to this cover. Note that $K_i\subseteq\operatorname{supp}\psi_i\subseteq\mathcal{O}_i$ and $\psi_i|_{K_i}=1$. Define the partition of unity $\lambda_i$ by
$$
\lambda_i(x)=\frac{\psi_i(x)}{\sum_{j\in I}\psi_j(x)}
$$
This is well defined (since $\sum_{i\in I}\psi_i(x)\ge 1$ and the sum is locally finite) and has all the desired properties.
