# Bootstraplemma 7.9 in Bredon

I am currently looking at the following reasoning bootstraplemma in Bredon:

Let $P_M (A)$ be a statement for the subset $A$ of the manifold $M$. Now, we have that the statement is true for all finite unions of compact $A \subset M$ contained in euclidean open sets. Then, he says that if (iii) "$P(A_i)$ holds for a decreasing chain of compact sets $\ldots \subset A_2 \subset A_1$, then $P(\cap_i A_i)$ holds" since any compact set is the intersection of finite unions as above (saying that we use that a compact set is separable metrizable). And this last step, I do not understand. Why is every compact set of such a form?

Let $K$ be your compact set. For every $n$, arrange a finite cover $\{B_1^n,\cdots,B^n_{i_n}\}$ of balls of radius less than $1/n$ for $K$, having centers in $K$. By the Lebesgue number lemma, there exists a $\delta>0$ such that any ball of radius $\delta$ centered in someone in $K$ is contained in some chart. Let $N$ be such that $1/N < \delta/2$.
Letting $A_n:=\bigcup\limits_{j=1,\cdots,i_n} \overline{B_j}^n,$ we will show that $K=\bigcap\limits_{j \geq N} A_j$.
First, note that $K \subset \bigcap\limits_{j \geq N} A_j$ is obvious. Now, pick $x \in \bigcap\limits_{j \geq N} A_j$.
For every $n \geq N$, there exists then a $x_n \in K$ such that $d(x,x_n) \leq 1/n$. It follows that $x$ is a limit point of $K$. Since $K$ is a compact in a Hausdorff space, it is closed, and hence $x \in K$.
EDIT: I've realized that the intersection must be of a decreasing chain, and the $A_i$'s as constructed above may not be decreasing. But you can adjust them to be by intersecting each one with the sets of the previous and using the "identity" mentioned by Bredon, i.e., $$A\cap (B_1 \cup \cdots \cup B_k)=(A \cap B_1) \cup \cdots \cup (A \cap B_k),$$ to show that taking such intersections do not mess with the property that each $A_i$ must be a finite union of "small" compact sets.