Let $U \subseteq \mathbb{R}^n$ be open. By using balls with rational centers and rational radii, we can get a countable collection $B_n$ of open balls so that $\bigcup B_n = U$.

I am also familiar with the fact that $\mathbb{R}^n$ (in fact any metrizable space) is paracompact. Applied to this particular situation, it means that we can find an open cover $\mathcal{B}$ of $U$ so that each $V \in \mathcal{B}$ is contained in some $B_n$, and each $a \in U$ has a neighborhood that meets only finitely many elements of $\mathcal{B}$.

My question is, is a there a construction one can perform so that the elements of $\mathcal{B}$ are very nice open sets? So nice, in fact, that $\mathcal{B}$ is countable and is just a set of open balls? It seems that, since $\{B_n \}$ started out as a countable collection of balls, $\mathcal{B}$ could end up that way as well. But we need more that just the fact that $U$ is paracompact for this to work out.

I will admit that I have not gotten very far with any potential construction. So maybe just some hints to start me out? Perhaps I will need to write this open set $U$ as an increasing sequence of compact sets?

Thanks in advance.

  • $\begingroup$ "Perhaps I will need to write this open set U as an increasing sequence of compact sets?" I'm not sure if that is necessary to obtain the result, but it would make it easy to obtain. $\endgroup$ Commented Sep 1, 2013 at 23:00

1 Answer 1


Let $U \subset \mathbb{R}^n$ open, and $\mathfrak{V} = \{V_\alpha : \alpha \in A\}$ an open cover of $U$. Let $\varnothing = K_0 \neq K_1 \subset K_2 \subset \dotsc$ a normal exhaustion of $U$, that is, all $K_n$ are compact, $K_n \subsetneq \overset{\circ}{K}_{n+1}$, and $U = \bigcup\limits_{n=0}^\infty K_n$. Then we get a locally finite refinement of the cover $\mathfrak{V}$ consisting of balls as follows:

Let $\mathcal{B}_0 = \varnothing$.

Let $\mathfrak{B}_1$ the set of all (nonempty) open balls (with rational centre and radius, if desired) that are contained in $\overset{\circ}{K}_2$ and in some $V_\alpha$. Cover $K_1$ with finitely many balls in $\mathfrak{B}_1$. Let $\mathcal{B}_1$ the set and $U_1$ the union of these balls.

For $n > 1$, if $K_{n-1}$ is already covered, let $\mathfrak{B}_n$ the set of all open balls (with rational centre and radius, if desired) that are contained in $\overset{\circ}{K}_{n+1} \setminus K_{n-1}$ and in some $V_\alpha$. Since the open balls with rational centre and radius form a basis of the topology, the compact set $K_n \setminus U_{n-1}$ is covered by such balls; select a finite subcover $\mathcal{B}_n$, and let $U_n = U_{n-1} \cup \bigcup \mathcal{B}_n$.

Inductively, we obtain a countable cover $\mathcal{B} = \bigcup\limits_{n=1}^\infty \mathcal{B}_n$ of $U$ by open balls that refines $\mathfrak{V}$.

$\mathcal{B}$ is locally finite: Let $x\in U$. Then $x \in K_n \setminus K_{n-1}$ for some $n > 0$. The open neighbourhood $\overset{\circ}{K}_{n+1} \setminus K_{n-1}$ of $x$ intersects only balls from $\mathcal{B}_k$ for $k\in \{n-1, n, n+1\}$, and that is a finite set.

  • $\begingroup$ Thank you very much for the answer, very clear! $\endgroup$
    – JZS
    Commented Sep 2, 2013 at 15:12

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