# Show there is a countable collection of closed balls $\beta=\{B_n=\overline{B_{r_n}(x_n)}:n\in \mathbb{N}\}$ such that $U=\cup_nB_n^{\text{int}}$

Let $$U= \cup C$$ be an open set in $$\mathbb{R}^n$$. Show there is a countable number of closed balls $$\beta=\{B_n=\overline{B_{r_n}(x_n)}:n\in \mathbb{N}\}$$ such that:

a.$$U=\cup_n B_n^{\text{int}}$$

b. For each $$n\in \mathbb{N}$$ there exists an open set $$M$$ such that $$B_n\subseteq M$$.

c. For every point $$x\in U$$, there exists a neighbourhood $$V$$ of $$x$$ such that only finitely many elements of the cover $$C$$ intersect $$V$$ non trivially.

So I want to use compact exhaustion of the open set $$U$$.

Fix a compact exhaustion $$K_n:n\in\mathbb{N}$$ of $$U$$

Then since by definition for every $$i\in\mathbb{N}$$ $$K_i\subseteq K_{i+1}^{\text{int}}$$

I can make a compact set $$L_n=K_n\setminus K_{n-1}^{\text{int}}$$ which will satisfy that each $$L_i$$ is compact as its an intersection of closed sets, which will be closed and is bounded by $$K_i$$.

From here I think what I want to do is to take balls $$B_i(x)$$ for each $$x\in L_i$$ and take a radius which will not intersect more then $$L_{i+1}$$ or $$L_{i-1}$$. A set of all such balls $$B_n$$ in each $$L_i$$ will admit a finite subcover because each $$L_i$$ is compact. And the union of all such finite subcovers of each $$L_i$$ should still cover $$U$$ and because $$L_n$$ is countable this set $$\cup B_n$$ will be countable.

I believe this would satisfy part a. but I don't really see part b and c. I also don't know what a non trivial intersection is.

• I think you can do this by considering sets of dyadic cubes, and the corresponding inscribed/circumscribed balls. – Matematleta Mar 29 at 22:52
• I've never seen dyadic cubes before. – AColoredReptile Mar 29 at 22:52
• Is there an extra part of the condition in b), or what prevents us to take $M=\mathbb{R}^n$ or $M=U$ for all $n$? – Henno Brandsma Mar 30 at 13:10
• non-trivial intersection is onon-empty intersection, I think. – Henno Brandsma Mar 30 at 13:11
• What is the cover $C$ in condition c)? Do you mean $\beta$? – Paul Frost Mar 30 at 17:24

Here's an answer to part a: This is just the same as, or a slight modification of the proof that was posted earlier.

For any set $$U$$ the set of balls $$\mathcal{B}=\{B_p(q)^{\mathrm{int}}~|~q\in \mathbb{Q}^n, p \in \mathbb{Q}_{\ge 0}\}$$ form a countable cover of $$U$$.

Let $$x$$ be an arbitrary point in $$U$$. Then $$x$$ is contained in an open ball $$B_{r_x}(x) \subseteq U$$. Let $$q$$ be a rational point contained within the ball $$B_{r_{x/3}}(x)$$. Then we can choose a ball $$B_p(q) \in \mathcal{B}$$ with $$r_x/3 < p <2r_x/3$$. Then $$x \in B_p(q) \subset B_{r_x}(x) \subseteq U.$$

Hence $$U$$ is the union of a subset of $$\mathcal{B}$$.

Take a family $$V=\{V_n:n\in \Bbb N\}$$ of bounded open sets such that $$U=\cup V$$ and such that $$\overline V_n \subset V_{n+1}$$ for each $$n.$$ (To get $$V,$$ see APPENDIX below.)

Let $$K_1=\overline V_1.$$ For $$n>1$$ let $$K_n=\bar V_n \ V_{n-1}.$$ Each $$K_n$$ is compact.

Let $$C_1$$ be a finite cover of $$K_1$$ by open balls, such that $$\overline {\cup C_1}\subset V_2.$$ This is possible because $$K_1$$ is compact, $$V_2$$ is open, and $$K_1\subset V_2.$$

For $$n>1$$ let $$C_n$$ be a finite cover of $$K_n$$ by open balls, such that $$\overline {\cup C_n}\subset V_{n+1}$$ and also such that $$\cup C_n$$ is disjoint from $$\cup_{j\le n-2}(\overline {\cup C_j}).$$

The last condition above is vacuous when $$n=2.$$ When $$n>2$$ it is crucial, and possible because $$S_n=\cup_{j\le n-2}(\overline {\cup C_j})$$ is a subset of $$\cup_{j\le n-1}V_j=V_{n-1},$$ and $$V_{n-1}$$ is disjoint from $$K_n,$$ so the closed set $$S_n$$ and the compact set $$K_n$$ are disjoint.

Now $$C=\cup_{n\in \Bbb n}C_n$$ is a countable family of open balls, and $$U=\cup C=\cup _{b\in C}\, int(\bar b)=\cup_{b\in C}\,\bar b.$$

If $$p\in U$$ then for some $$n$$ we have $$p\in V_n\subset \overline V_n=\cup_{j\le n}K_j$$ so for some $$j\le n$$ and some $$b \in C_j$$ we will have $$p\in b\in C_j.$$ Now this $$b$$ is disjoint from $$\cup C_m$$ for every $$m\ge n+2.$$ So $$\{b'\in C: b'\cap b\ne \phi\}$$ is a subset of the finite set $$\cup_{j\le n+1}C_j.$$

APPENDIX. Let $$\bar 0$$ be the origin in $$\Bbb R^n.$$

(i). If $$U=\Bbb R^n$$ let $$V_n=B_n(\bar 0)$$ for each $$n\in \Bbb N.$$

(ii). If $$U\ne \Bbb R^n,$$ let $$U^c=\Bbb R^n$$ \ $$U$$ and let $$d$$ be a metric for the topology on $$\Bbb R^n,$$ and for $$p\in U$$ let $$d(p,U^c)=\inf \{d(p,q):q\in U^c\}.$$

Now let $$V_1=\{p\in U\cap B_1(\bar 0): d(p,U^c)>1\}.$$ And for $$n>1$$ let $$V_n= \{p\in U\cap B_n(\bar 0): d(p,U^c)>1/n\}.$$ I will leave it to the reader to confirm that $$\{V_n:n\in \Bbb N\}$$ has the required properties.

• This works if you replace $\Bbb R^n$ with any metric space in which closed bounded subsets are compact. – DanielWainfleet Mar 31 at 18:37