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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....

Thank you in advance!

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1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – J. C.
    Dec 24, 2022 at 17:50
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    $\begingroup$ For future reference: my question in the above comment was answered here. $\endgroup$
    – J. C.
    Jan 8, 2023 at 18:03
  • $\begingroup$ @J.C. Thanks a lot! $\endgroup$
    – B.Hueber
    Jan 8, 2023 at 18:24

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