Spivak Calculus on Manifolds - Problem 1.22

Question

Problem 1.22 - Spivak's Calculus on manifolds

If $U$ is open $C \subset U$ is compact, show that there is a compact set $D$ such that $C \subset interior \ D$ and $D \subset U$.

observation: the metric space is $\mathbb{R}^n$.

My attempt:

$U$ is an open set, so exists $\delta_x > 0$ such that $B(x,\delta_x) \subset U$ for each $x \in U$, therefore the collection $\{ B(x,\delta_x) \}$ is an open cover for $C$ since $U = \bigcup_{x \in U} B(x,\delta_x)$ and $C \subset U$ then exists a finite subcollection $\{ B(x_i,\delta_{x_i}) \ ; \ i = 1, \cdots, n \}$ that covers $C$ by compactness of $C$. I suppose that my candidate of $D$ is $\overline{ \bigcup_{i = 1}^{i = n} B(x_i,\delta_{x_i}) }$ since the set $\bigcup_{i = 1}^{i = n} B(x_i,\delta_{x_i})$ is open and $interior \ D = \bigcup_{i = 1}^{i = n} B(x_i,\delta_{x_i})$ and $C \subset \ interior \ D$.

I would like to know if I'm right of thinking that $D = \overline{ \bigcup_{i = 1}^{i = n} B(x_i,\delta_{x_i}) }$ and I would like to know how to prove that $D$ is compact and that $D \subset U$.

Thanks in advance!

Answer 1

You are almost there, just replace $B(x_i,\delta_{x_i})$ by $B(x_i,\delta_{x_i}/2)$ and take $D=\bigcup_{i=1}^{i=n}\overline{B(x_i,\delta_{x_i}/2)}$.

$\overline{B(x_i,\delta_{x_i}/2)}\subset B(x_i,\delta_{x_i}) $ implies $D\subset U$. $D$ is compact since a closed ball in $R^n$ is compact and a finite union of compact subsets is compact since it is closed and bounded.

Answer 2

In the first line, you claim that there exists $d > 0$ such that $B(x,d)$ is contained in $U$ for all $x$. However, that may not be the case, as the d depends on x. However, if you use $d_x$ and $d_x/2$ in the proof then this doesn't really affect anything.