# Is it always possible to find a partition of unity subordinate to a cover with the same index set?

Let $$(X,\mathcal{T})$$ be a topological space. A parition of unity subordinate to an open cover $$(\mathcal{O}_{i})_{i\in I}\in\mathcal{T}^{I}$$ is a collection of maps $$\{f_{j}:X\to [0,1]\}_{j\in J}$$ such that

1. The set of supports $$\{\operatorname{supp}(f_{j})\}_{j\in J}$$ is locally finite, which means that every point has a neighbourhood, which intersects only finitely many elements of $$\{\operatorname{supp}(f_{j})\}_{j\in J}$$.
2. For every $$j\in J$$ there is an $$i\in I$$ such that $$\operatorname{supp}(f_{j})\subset U_{i}$$.
3. $$\forall x\in X:\sum_{j\in J}f_{j}(x)=1$$

Often, we are interested in a partition of unity $$\{f_{i}:X\to [0,1]\}_{i\in I}$$ subordinate to a cover $$(\mathcal{O}_{i})_{i\in I}\in\mathcal{T}^{I}$$ with the same index set such that $$\forall i\in I:\operatorname{supp}(f_{i})\subset U_{i}$$.

If there exists an partition of unity subordinate to a cover, can we always choose without loss of generality that it has the same index set?

I was thinking of the following proof:

Proof: Let $$\{f_{j}:X\to [0,1]\}_{j\in J}$$ be a subordinate partition of unity subordinate to an open cover $$(U_{i})_{i\in I}$$. Then there is for every $$j\in J$$ an $$i\in I$$, such that $$\operatorname{supp}(f_{j})\subset U_{i}$$. Let $$\varphi:J\to I$$ be the map which sends every $$j\in J$$ to the corresponding $$i\in I$$. We define for every $$i\in\varphi(J)$$ the map $$\widetilde{f}_{i}:X\to [0,1]$$ for all $$x\in X$$ through \begin{align*}\widetilde{f}_{i}(x):=\sum_{j\in\varphi^{-1}(\{i\})}f_{j}(x)\end{align*} and for every $$i\in I$$ \ $$\varphi(J)$$, $$\widetilde{f}_{i}$$ to be the constant zero function. Then is $$\{\widetilde{f}_{i}:X\to [0,1]\}_{i\in I}$$ obviously a partition of unity subordinate to an open cover $$(U_{i})_{i\in I}$$ with $$\forall i\in I:\operatorname{supp}(\widetilde{f}_{i})\subset U_{i}$$. $$\blacksquare$$

But the problem is, this works only, if $$\varphi^{-1}(\{i\})$$ is finite for all $$i\in\varphi(J)$$, or in other words, if every set $$U_{i}$$ of the cover contains only finitely many supports. Otherwise, the sum is not well-defined......So the question is, if this is true? Maybe this has to do something with the locally finiteness of the supports....

Oh I think I had found an answer by myself....$$\varphi^{-1}(\{i\})$$ isn't finite, but by locally finiteness of $$\{\operatorname{supp}(f_{j})\}_{j\in J}$$, we know that only finitely many $$f_{j}(x)$$ with $$j\in \varphi^{-1}(\{i\})$$ are non-zero, because every $$x\in X$$ (and therefore also every $$x\in U_{i}$$) has a neighbourhood, which intersects only finitely many of the supports. Following this, the sum is well-defined.
• Even if the sum is finite for each $x$, how do we know that the sum still has support contained in $U_i$? If this open set contains infinitely many supports then it seems to me that it would be possible that the sum is nonzero in the entire open set. Dec 24, 2022 at 17:50