I am reading "Calculus on Manifolds" by Michael Spivak.
I am now reading the case 1 in the proof of Theorem 3-11 about partitions of unity.
Unfortunately I could not understand the case 1 in the proof of Theorem 3-11.
I found the following question.
Unfortunately I counld not understand what the answers say.
Partition of Unity in Spivak's Calculus on Manifolds
But yesterday I thought maybe I finally understood the case 1 in the proof of Theorem 3-11.
However, I am not confident that I understand correctly.
I want you to check if I understand the case 1 in the proof of Theorem 3-11.
My summary of Case 1 in the proof of Theorem 3-11:
Let $A$ be a compact subset of $\mathbb{R}^n$.
Let $\mathcal{O}$ be an open cover of $A$.
Since $A$ is compact, $\{U_1,\dots,U_n\}\subset\mathcal{O}$ covers $A$.
It clearly suffices to construct a partition of unity subordinate to the cover $\{U_1,\dots,U_n\}$.
The author constructed the compact sets $D_1,\dots,D_n$ such that $D_1\subset U_1,\dots,D_n\subset U_n$ and {$\text{ int }D_1,\dots,\text{ int }D_n\}$ is an open cover of $A$.
Then, the author used the result of Problem 2-26(d):
Problem 2-26(d)
If $A\subset\mathbb{R}^n$ is open and $C\subset A$ is compact, show that there is a non-negative $C^\infty$ function $f:A\to\mathbb{R}$ such that $f(x)>0$ for $x\in C$ and $f=0$ outside of some closed set contained in $A$.
Problem 2-26(e)
Show that we can choose such an $f$ so that $f:A\to [0,1]$ and $f(x)=1$ for $x\in C$.
By Problem 2-26(d), for each $i\in\{1,\dots,n\}$, there is a non-negative function $\psi_i:U_i\to\mathbb{R}$ such that $\psi_i(x)>0$ for $x\in D_i$ and $\psi_i=0$ outside of some closed set contained in $U_i$.
Since $\{D_1,\dots,D_n\}$ covers $A$, we have $\psi_i(x)+\cdots+\psi_n(x)>0$ for all $x$ in some open set $U$ such that $A\subset U\subset \text{ int }(D_1\cup\dots\cup D_n)\subset D_1\cup\dots\cup D_n$.
Let $\varphi_i:U\to\mathbb{R}$ be the function such that $\varphi_i(x)=\frac{\psi_i(x)}{\psi_1(x)+\cdots+\psi_n(x)}$ for each $i\in\{1,\dots,n\}$.
Then, the author used the result of Problem 2-26(e) above.
By Problem 2-26(e), there is a function so that $f:U\to [0,1]$ and $f(x)=1$ for $x\in A$ and $0$ outside of some closed set contained in $U$.
Then, $\Phi=\{f\cdot\varphi_1,\dots,f\cdot\varphi_n\}$ is the desired partition of unity.
I could not understand this proof.
(4) says the following:
(4) For each $\varphi\in\Phi$ there is an open set $U$ in $\mathcal{O}$ such that $\varphi=0$ outside of some closed set contained in $U$.
I thought $U$ (which is the domain of the functions in $\Phi$) is too small.
I thought it is possible that there is an open set $U_i\in\mathcal{O}$ such that $U_i-U\neq\varnothing$.
And I could not understand why $\Phi$ satisfies (4).
Yesterday, I thought maybe I finally understood the case 1 in the proof of Theorem 3-11.
I modified the last part of the case 1 in the proof of Theorem 3-11 to extend $U$ to sufficiently large open set as follows:
Let $A$ be a compact subset of $\mathbb{R}^n$.
Let $\mathcal{O}$ be an open cover of $A$.
Since $A$ is compact, $\{U_1,\dots,U_n\}\subset\mathcal{O}$ covers $A$.
It clearly suffices to construct a partition of unity subordinate to the cover $\{U_1,\dots,U_n\}$.
The author constructed the compact sets $D_1,\dots,D_n$ such that $D_1\subset U_1,\dots,D_n\subset U_n$ and {$\text{ int }D_1,\dots,\text{ int }D_n\}$ is an open cover of $A$.
Then, the author used the result of Problem 2-26(d):
Problem 2-26(d)
If $A\subset\mathbb{R}^n$ is open and $C\subset A$ is compact, show that there is a non-negative $C^\infty$ function $f:A\to\mathbb{R}$ such that $f(x)>0$ for $x\in C$ and $f=0$ outside of some closed set contained in $A$.
Problem 2-26(e)
Show that we can choose such an $f$ so that $f:A\to [0,1]$ and $f(x)=1$ for $x\in C$.
By Problem 2-26(d), for each $i\in\{1,\dots,n\}$, there is a non-negative function $\psi_i:U_i\to\mathbb{R}$ such that $\psi_i(x)>0$ for $x\in D_i$ and $\psi_i=0$ outside of some closed set contained in $U_i$.
Since $\{D_1,\dots,D_n\}$ covers $A$, we have $\psi_i(x)+\cdots+\psi_n(x)>0$ for all $x$ in some open set $U$ such that $A\subset U\subset \text{ int }(D_1\cup\dots\cup D_n)\subset D_1\cup\dots\cup D_n$.
Let $\varphi_i:U\to\mathbb{R}$ be the function such that $\varphi_i(x)=\frac{\psi_i(x)}{\psi_1(x)+\cdots+\psi_n(x)}$ for each $i\in\{1,\dots,n\}$.
Then, the author used the result of Problem 2-26(e) above.
By Problem 2-26(e), there is a function so that $f:U\to [0,1]$ and $f(x)=1$ for $x\in A$ and $0$ outside of some closed set $C$ contained in $U$.
$f\cdot\varphi_i:U\to\mathbb{R}$ is a $C^\infty$ function such that $f\cdot\varphi_i(x)=\varphi_i(x)$ for $x\in A$ and $0$ outside of $C$ which is contained in $U$ for each $i\in\{1,\dots,n\}$.
We can extend the domain of $f\cdot\varphi_i$ from $U$ to $\mathbb{R}^n$ as follows:
$(\text{ new }f\cdot\varphi_i)(x):=0$ if $x\in\mathbb{R}^n-U$.
$(\text{ new }f\cdot\varphi_i)(x):=(\text{ old }f\cdot\varphi_i)(x)$ if $x\in U$.
This $\text{ new }f\cdot\varphi_i$ is also a $C^\infty$ function.
Let $\text{ new } U:=\mathbb{R}
^n$.
Then $\Phi=\{f\cdot\varphi_1,\dots,f\cdot\varphi_n\}$ whose elements are defined on $\text{ new }U$ is the desired partition of unity:
(1) holds because for each $x\in A$ we have $0\leq f\cdot\varphi_i(x)=\varphi_i(x)\leq 1$.
(2) obviously holds because $\Phi$ is a finite set.
(3) holds because for each $x\in A$ $\sum_{\varphi\in\Phi}\varphi(x)=\sum_{i=1}^n f\cdot\varphi_i(x)=\sum_{i=1}^n \varphi_i(x)=\sum_{i=1}^n \frac{\psi_i(x)}{\psi_1(x)+\cdots+\psi_n(x)}=1$.
(4) holds:
Let $\varphi=f\cdot\varphi_i\in\Phi$.
Then, $\varphi=0$ outside of $C\cap D_i$ which is closed (compact) set containd in $U_i\in\mathcal{O}$.
