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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.

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  • $\begingroup$ I think you can do this by considering sets of dyadic cubes, and the corresponding inscribed/circumscribed balls. $\endgroup$ – Matematleta Mar 29 at 22:52
  • $\begingroup$ I've never seen dyadic cubes before. $\endgroup$ – AColoredReptile Mar 29 at 22:52
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    $\begingroup$ 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 $n$? $\endgroup$ – Henno Brandsma Mar 30 at 13:10
  • $\begingroup$ non-trivial intersection is onon-empty intersection, I think. $\endgroup$ – Henno Brandsma Mar 30 at 13:11
  • $\begingroup$ What is the cover $C$ in condition c)? Do you mean $\beta$? $\endgroup$ – Paul Frost Mar 30 at 17:24
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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 $\mathcal{B}$.

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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.

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

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