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Let $A\subseteq \mathbb{R}^n$ and consider a family of open balls $(B(x_i,r_i))_i$ indexed on a set $I$ such that $$A\subseteq \cup_{i\in I}B(x_i,r_i)$$

Prove that there exists an (at most) countable subset $I*\subset I$ such that $$A\subseteq \cup_{i\in I*}B(x_i,r_i)$$

My thoughts on the matter: First we need an ordering on I. This is always possible (?). We begin by a point $a\in A$ and if $a \in B(x_{i_1},r_{i_1})$, then either $A\subset B(x_{i_1},r_{i_1})$ and we are done with $I={i_1}$, or we can choose a point $a_2\in A_1=A-\cup_{i<i_1}B(x_i,r_i)$. This construction would create a sequence of indexes $i_1<i_2<\ldots$ with the desired property, if somehow we could also exhaust the index set I. This is not clear at all though...

Is there a better approach?

If I, say, let $I=\{i:i<\aleph_0\}\cup\{i:i<\aleph_1\}\ldots$ would this imply that $I$ is exhausted?

MelaniesWoes
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1 Answers1

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Let $I$ be a index set, $\{\mathcal{O}_i \}_{i\in I}$ be a family of open sets in $\mathbb{R}^n$. If $$A \subset \bigcup_{i\in I} \mathcal{O}_i := \mathcal{O} $$ we have to prove that there is a countable subcovering.

Since $\mathcal{O}$ is open, it is $F_\sigma$, so there are closed sets $K_1, K_2, K_3 \cdots$ whose union is $\mathcal{O}$. By intersecting each $K_i$ with closed ball of finite radius, we may assume all $K_i$ are compact. Each $K_n$ has a finite subcovering from $\{\mathcal{O}_i \}$. Thus we have obtained countable many $\mathcal{O}_i$ covering $\mathcal{O} \supset A$. This completes the proof.

pisco
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  • Good approach. But, in the bounded case, how do we know if $U=\cup_{x\in A}U_x$ is countable? Don't we need something more? – MelaniesWoes Dec 26 '17 at 13:45
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    @MelaniesWoes See my edit. Btw, all we need to know is $U$ is open, hence $F_\sigma$, which is a countable union of closed set. – pisco Dec 26 '17 at 13:50