# How does partition of unity imply that for a compact $D$, there is only finitely many $\operatorname{Support}\phi_i$ that intersect with $D$?

In the book of Analysis On Manifolds by Munkres, at page 141, it is given that

However, I think there is a problem in the existence such an $M$.

Reason:

Because we do know that the supports of $\phi_i$s, i.e $S_i$s, will form a cover for $A$, and hence for $D$, but by the compactness of $D$, we can find a finitely many $S_i$s that will cover $D$, but this does not mean that $D$ has a non-empty intersection with only finitely many $S_i$s. Therefore, there might be cases where $D$ have intersect with infinitely many $S_i$s, so the existence of $M$ is not clear, IMO.

I should point out also that the very definition of $\phi$s requires that for any $x\in A$, there exists an neighbourhood of $x$ s.t only finitely many of $S_i$ will intersect with that nbd.However, $D$ possibly have infinitely many element, so I can't see how can one derive that from the local finiteness the existence of $M$.

We assume the partition of $1$ to be locally finite, i.e. for each $x\in A$ there is an open neighbourhood $U_x$ of $x$ and a number $M_x$ with $\phi_i(y) = 0$ whenever $y\in U_x$ and $i>M_x$.
$D$ is covered by the collection $\{U_x\vert x \in D\}$ and since it is compact, there is a finite subset $F\subset D$ with $D\subset\bigcup_{x\in F} U_x$. Then $M=\max_{x\in F} M_x$ does the job: Let $y\in D$ and $i>M$. Then there is an $x\in F$ with $y\in U_x$ and since $i>M\ge M_x$ we have $\phi_i(y)=0$.