This question is strongly related to this one, but the existing answer doesn't cover the point I will ask about below.
The context I am concerned about is the following. Let $S \subseteq \mathbb{R}^n$ be some subset, and $\{U_i\}$ some open cover of $S$. A smooth partition of unity for $S$ subordinate to the open cover $\{U_i\}$ is a collection $\{\varphi_j\}$ of smooth (i.e. $C^\infty$) functions $\varphi_j: V \to [0,1]$ defined on some open neighbourhood $V$ of $S$, such that:
- For each $j$, there exists some $i$ such that $\operatorname{supp} \varphi_j \subseteq U_i$;
- Each $x \in S$ has an open neighbourhood where all but finitely many of the $\varphi_j$ vanish;
- For all $x \in S$, $\sum_j \varphi_j(x) = 1$ (this is a finite sum by point 2).
The existence of such partitions of unity is proved, for example, in Spivak's Calculus on Manifolds (Theorem 3-11).
However, sometimes we like that the partition of unity is indexed by the same set as the open cover.
Is it true that we can always modify a partition of unity as defined above such that it has the same index set as the open cover?
The first approach one would think of trying is something akin to the one in the linked question: choose a map from the $j$ indexes to the $i$ indexes (respecting condition 1 above) and then sum all the $\varphi_j$ corresponding to the same $i$. The linked question was concerned about the well-definedness of this sum (it is well-defined, since it is actually a finite sum for each $x$, and with a bit more care we can see that this gives a smooth function, at least if we construct the partition of unity following Spivak's proof). However, we still need that this sum has support contained in $U_i$. This is clear in the case where we only have finitely many $j$ corresponding to the same $i$, but in the infinite case it seems to me that the sum could possibly be nonzero over the whole open $U_i$.
I have a feeling that maybe this difficulty could be avoided by some clever multiplication with a cutoff function which makes the $\varphi_j$ vanish close to the boundary in a careful way, but I couldn't actually make it work.
Best Answer
The problem you are worried about cannot occur. Indeed, condition 2. implies that the family $\{A_j:=\text{supp}(\varphi_j)\}$ is locally finite, i.e.~any point $x$ has an open neighborhood that intersects only finitely many members of the family. Hence the same is true for the subfamily consisting of those $\varphi_j$ for which the chosen index is some fixed $i_0$. Let us write this as $j\in J_0$. But then it is a fact of general topology that the union of a locally finite family of closed sets is closed. (For $x\notin \cup_{j\in J_0}A_j$, there is an open neighborhood $U$ of $x$ that intersects only finitely many $A_j$. But then the the intersection of the finitely many sets $U\setminus A_j$ is an open neighborhood of $X$ which is contained in the complement of $\cup_{j\in J_0}A_j$.) Now if you define $\psi:=\sum_{j\in J_0}\phi_j$ then this is smooth and $\{y:\psi(y)\neq 0\}\subset\cup_{j\in J_0}A_j$. But since the union is closed, you see that $\text{supp}(\psi)\subset \cup_{j\in J_0}A_j\subset U_i$, which is all you need.