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On Wikipedia a partition of unity is a collection of continuous maps $\varphi_i$ from a topological space $X$ into $\mathbb R$ such that for all $x$

(i) $\sum_i \varphi_i (x) = 1$

(ii) there is a neighbourhood $N_x$ such that only finitely many $\varphi_i$ are non-zero on $N_x$

The definition of partition of unity in Frank Warner's "Foundations of Differentiable Manifolds and Lie Groups" is given as

(i) $\sum_i \varphi_i (x) = 1$

(ii) there is a neighbourhood $N_x$ such that $N_x$ intersects only finitely many of the $\mathrm{supp} \varphi_i$

The support of a function is the closure of the set where the function is not zero.

Hence my question: Are these definitions equivalent? It appears to be the case that there are less points in each $N_x$ in the second definition because the closure also contains points where $\varphi_i$ is zero.

1 Answers1

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We only have to show that the condition in Wikipedia implies the one by Warner. Let

$$ M_i = \{x \mid \varphi_i(x) \neq 0\}. $$

By assumption, this is a locally finite family. We want to show that this is also true of the family $\overline{M_i}$.

This is true in general, because for each $x$, there is an (open) neighborhood $U$ with $M_i \cap U \neq \emptyset$ only for $I$ in some finite set. For other $I$, this yields $M_i \subset U^c$, which is closed. Hence $U \cap \overline{M_i} = \emptyset$ for all but (the same) finite set of indices $i$.

PhoemueX
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  • Thank you, very helpful! So basically it boils down to this: If an open set $U$ does not intersect a set $V$ then it also does not intersect $\overline{V}$. –  Oct 16 '14 at 05:50
  • Yes, exactly. Because this statement is so general, we did not even have to use any properties of the $\varphi_i$, like continuity. – PhoemueX Oct 16 '14 at 07:55