Can one or more element of a partition of unity be identically zero?

Question

Probably a silly or obvious question! But for the sake of clarity, let's take the open subset of $\mathbb{R}$ $(-1,1)$ and its open cover given by all the open sets $U_a = (-a,a)$ with $0<a<1$. Such open cover must admit a partition of unity subordinate to it $\sum_a{\rho_a}=1$, but in order for $\sum_a{\rho_a}$ to be a locally finite sum I guess only a subset of the $\rho_a$ elements can be different from zero. For instance, all identically zero but $\rho_1=1$ over $U$ may satisfy the requirement. Is that correct?

Answer 1

It's difficult to say anything with certainty, since definitions of partitions of unity vary somewhat. That said, it is generally allowed that some elements of the partition vanish, in the following sense:

Given an open cover $C=\{U_i\subseteq M:i\in I\}$ of a smooth manifold $M$, one can define a partition of unity subordinate to $C$ to be a collection $(\psi_i\in C^\infty M:i\in I)$ such that $\operatorname{supp}(\psi_i)\subset U_i$, for each $x\in M$ there is a neighborhood of $x$ on which all but finitely many $\psi_i$ vanish, and $\sum_i\psi_i=1$ (this sum being well defined due to local finiteness).

There is nothing in this definition that requires $\psi_i\neq 0$, and, in order for partitions to exist subordinate to any open cover, we must allow this case. For instance, every manifold is Lindelöf, so we can find a countable cover of neighborhoods on which $\psi_i$ is locally finite, which implies only countably many $\psi_i$ are nonvanishing, and thus some $\psi_i$ must vanish if the index set $I$ is uncountable. Additionally, all but finitely many $\psi_i$ must vanish on a compact manifold by a similar argument, and so any partition subordinate to an infinite cover of a compact space will likewise have $\psi_i=0$ for all but finitely many $\psi_i$.

Your choice of cover is one such example by the Lindelöf argument, though as noted in the comments the function $1$ is not supported by any element of the cover. One can instead note that, given a POU $\psi_a,a\in(0,1)$ subordinate to your cover, there is by local finiteness a neighborhood $(-\epsilon,\epsilon)$ of $0$ such on which all but finitely many $\psi_a$ vanish. This means that among the infinitely many $\psi_a$ with $a<\epsilon$, all but finitely many must be identically zero.