Is it really true that only finitely many $\phi \in \Phi$ are not $0$ on $C$? ("Calculus on Manifolds" by Michael Spivak)

Question

I am reading "Calculus on Manifolds" by Michael Spivak.

Let $C \subset A$ be compact.

Spivak wrote:

"For each $x \in C$ there is an open set $V_x$ containing $x$ such that only finitely many $\phi \in \Phi$ are not $0$ on $V_x$. Since $C$ is compact, finitely many such $V_x$ cover $C$. Thus only finitely many $\phi \in \Phi$ are not $0$ on $C$".

I think it is possible $\phi \in \Phi$ is $0$ on $V_x$, but not $0$ on $C \setminus V_x$.

Is it really true that only finitely many $\phi \in \Phi$ are not $0$ on $C$?

Answer 1

The proof of this fact is the following. For $x \in C$ let's denote by $V_x$ a neighborhood of $x$ such as there is only finitely many $\phi$ that are non zero on $V_x$

As $C$ is compact, we know that there is a finite number of points $(x_k)_{k=1}^n$ such that: $$ C \subset \bigcup_{k=1}^n V_{x_k}$$

Then because $\{\phi, \phi|_C \neq 0\} \subset \bigcup_{k=1}^n \{\phi, \phi|_{V_{x_k}} \neq 0\}$ (If you are non zero on $C$, then there is at least a $V_{x_k}$ where you are non zero) and because all the set on the right are finite, we can conclude that the set on the left is also finite.

Answer 2

Spivak's Theorem 3-11 states that there exists a collection of functions $\phi$ with suitable properties.

Of course it is possible that $\phi \in \Phi$ is $0$ on $V_x$, but not $0$ on $C \setminus V_x$ and thus not $0$ on $C$.

I think your doubts come from the fact that we may have infinitely many $\phi \in \Phi$ - so why shouldn't be infinitely many of them not $0$ on $C$?

So let us study Spivak's argument. $C$ is compact, hence it is covered by finitely many $V_{x_i}$ having the property that only finitely many $\phi_{i,j}$ are not $0$ on $V_{x_i}$. Now consider $\phi \notin \{ \phi_{i,j} \}$. Then $\phi$ is $0$ on all $V_{x_i}$, thus $0$ on $C$.