Spivak Calculus on Manifolds - Problem 2.26 (d)

Question

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 $f: A \longrightarrow \mathbb{R}$ such that $f(x) > 0$ for $x \in C$ and $f = 0$ outside of some closed set contained in $A$.

This is what I think at the moment:

$A$ is open, so there is $\delta_x > 0$ for each $x \in A$ such that $B(x,\delta_x) \subset A$, then $\{ B(x,\delta_x) \ ; \ x \in A \}$ is an open cover for $C$ and there is a finite subcollection $\{ B(x_i,\delta_{x_i}) \ ; \ i = 1, \cdots, n \}$ that covers $C$ by the compactness of $C$. Analogously, we obtain a finite subcollection $\{ B(x_i,\delta_{x_i}/2) \ ; \ i = 1, \cdots, n \}$ that covers $C$ by the compactness of $C$, so the set $D := \bigcup_{i=1}^{i=n} \overline{B(x_i,\delta_{x_i}/2)} \subset A$. I would like to define $f:A \longrightarrow \mathbb{R}$ such that $f(x) > 0$ for $x \in C$ and $f = 0$ for $x \in A - D$.

I would like to know if I'm the right way and to receive a hint about how construct this function. Thanks in advance!

Answer 1

I am now reading the proof of Theorem 3-11 in the section "PARTITIONS OF UNITY" in this book ("Calculus on Manifolds" by Michael Spivak).

And the author uses the result of Problem 2-26 in the proof of Theorem 3-11.

So I forced to solve Problem 2-26.

My solution:

I used the result of Problem 1-22 on p.10 in this book ("Calculus on Manifolds" by Michael Spivak).

Problem 1-22. If $U$ is open and $C\subset U$ is compact, show that there is a compact set $D$ such that $C\subset\text{interior }D$ and $D\subset U$.

Let $D$ be a closed (it is not necessary to be compact) set such that $C\subset\text{interior }D$ and $D\subset A$.

For each $x\in C\subset\text{interior }D$, there is an open rectangle $I_x$ such that $x\in I_x\subset \text{interior }D$ since $\text{interior }D$ is open.

Since $C$ is compact there is a collection $\{I_{x_1},\dots,I_{x_m}\}$ such that $C\subset I_{x_1}\cup\cdots\cup I_{x_m}\subset\text{interior }D$.

Let $B$ be the closure of $I_{x_1}\cup\cdots\cup I_{x_m}$.
Then $B\subset D\subset A$.

Let $g_{x_i}:\mathbb{R}^n\to\mathbb{R}$ be a $C^\infty$ function which is positive on $I_{x_i}$ and zero elsewhere for each $i\in\{1,\dots,m\}$. (such $g_{x_i}$ exists by the result of Problem 2-26(c) on p.29 in this book ("Calculus on Manifolds" by Michael Spivak)).

Let $f=g_{x_1}+\cdots+g_{x_m}$.

Then, $f$ is a $C^\infty$ function and $f(x)>0$ for $x\in C$ and $f=0$ outside of $B$ which is a closed set contained in $A$.

Answer 2

You may construct your closed set, i.e. $D$, at the start, but ultimately the proof comes down to (a) $C$ is compact and (b) is contained in an open set. We summarize the proof to three parts.

(i) If $x = (x_1, \ldots, x_n) \in C$, then $x \in U_1 \times \cdots U_n \subseteq A$, since $A$ is open, where $U_i$ are open intervals. This is by definition of product topology of $\mathbb{R}^n$. By taking subsets, $U_i$ are bounded intervals.

(ii) Use compactness to obtain a finite collection of open rectangles, $\{R_i \}$. We may define functions $g_i: \mathbb{R}^n \rightarrow \mathbb{R}$ such that $g_i \in C^{\infty}$ is positive on $R_i$ and $0$ elsewhere. Take the sum of all $g_i$.

(iii) Existence of a closed set: We can either replace step (i) with $A = \mathring{D}$ (your strategy), OR, choose sufficiently small open intervals in (i) such that $\bigcup_{ i=1}^n \overline{R_i}$ is contained in $A$.

Answer 3

$\newcommand{\Reals}{\mathbf{R}}$Here's a sketch based on rectangles instead of balls, for which the functions may be easier to construct:

If $x = (x_{1}, \dots, x_{n}) \in \Reals^{n}$ and if $\delta > 0$, let $$ R(x, \delta) = [x_{1} - \delta, x_{1} + \delta] \times \cdots \times [x_{n} - \delta, x_{n} + \delta] $$ be the axis-oriented cube of "radius" $\delta$ centered at $x$.

For each point $x$ and each $\delta > 0$, consider the "apparatus" $R(x, \delta/2) \subset R(x, \delta)$. It's straightforward to show that there exists a smooth, non-negative function $f_{x}$ that is identically $1$ on $R(x, \delta/2)$ and identically $0$ outside $R(x, \delta)$. (Define such a function in one dimension, then take the obvious product of one-variable functions.)

Now, for each $x$ in $C$, choose an apparatus such that the large box $R(x, \delta) \subset A$, then use compactness of $C$ together with the fact that the small boxes cover $C$ to pick finitely many apparatuses whose small boxes cover $C$. The sum of the corresponding functions $f_{x}$ is smooth, non-negative, positive on $C$, and identically $0$ outside $A$.