# Construction of partitions of unity in Warner

On p. 11 of Warner's Foundations of Differntiable Manifolds and Lie Groups, he discusses partitions of unity. The theorem says

Let $M$ be a differentiable manifold and $\{U_\alpha: \alpha \in A \}$ an open cover of $M$. Then there exists a countable partition of unity $\{ \varphi_i : i = 1, 2, 3, \dotsc \}$ subordinate to the cover $\{U_\alpha\}$ with $\operatorname{supp}\varphi_i$ compact for each $i$. If one does not require compact supports, then there is a partition of unity $\{\varphi_\alpha \}$ subordinate to the cover $\{U_\alpha \}$ (that is, $\operatorname{supp} \varphi_\alpha \subset U_\alpha$) with at most countably many of the $\varphi_\alpha$ not identically zero.

I am fine with the first part of the theorem: he constructs a countable partition of unity $\varphi_i$ with compact supports. Then, for the second part, he says

If we let $\varphi_\alpha$ be identically zero if no $\varphi_i$ has support in $U_\alpha$, and otherwise let $\varphi_\alpha$ be the sum of the $\varphi_i$ with support in $U_\alpha$, then $\{\varphi_\alpha\}$ is a partition of unity subordinate to the cover $\{U_\alpha\}$ with at most countably many of the $\varphi_\alpha$ not identically zero. To see that the support of $\varphi_\alpha$ lies in $U_\alpha$, observe that if $\mathcal{A}$ is a locally finite family of closed sets, then $\overline{\bigcup_{A \in \mathcal{A}}A} = \bigcup_{A \in \mathcal{A}} A$. Observe, however, that the support of $\varphi_\alpha$ is not necessarily compact.

If $\sum_\alpha \varphi_\alpha$ is to be identically $1$, then we have to make sure that we're not counting any $\varphi_i$ twice, right? But we will count it twice if $\operatorname{supp}\varphi_i$ lies in $U_\alpha$ for more than one $\alpha$, right? So perhaps we have to renormalize afterward...that would be OK, but how do we know that $\operatorname{supp}\varphi_i$ doesn't lie in infinitely many $U_\alpha$?

Does it have to do perhaps with the particular construction of the $\varphi_i$ he's chosen? (I don't see how.)

For instance, a stupid example which seems to break this construction is the open cover of a manifold $M$ by infinitely many copies of the single set $M$. Then take the function $\varphi$ which is identically 1 on $M$...this works for the first part. But in defining $\varphi_\alpha$ in the second part, we seem to be trying to sum up infinitely many copies of $\varphi$.

I think you're right. A better way to define $\varphi_\alpha$ would be to choose a map $a\colon \mathbb Z^+\to A$ such that $\operatorname{supp}\varphi_i\subseteq U_{a(i)}$ for each $i$, and then define $\varphi_\alpha$ to be $$\varphi_\alpha = \sum_{i: a(i)=\alpha} \varphi_i.$$ This is in fact the definition I used in my Introduction to Smooth Manifolds (2nd ed., p. 44).