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I'm trying to show:

$A\subset \mathbb{R}^n$ is open if and only if $A$ is the union of a collection of open balls accounting.

"$\Leftarrow$" Let $A=\bigcup_{i=1}^{\infty} \mathbb{B_{\varepsilon_i} (x_i)}$

If $y\in A $ then $y\in \mathbb{B}_{\varepsilon_i}(x_i)$ (for some $i$)

But note that $\mathbb{B}_{||x-y||}(y)\subset \mathbb{B}_{\varepsilon_i}(x_i)$ or $\mathbb{B}_{\varepsilon_i-||x-y||}(y)\subset \mathbb{B}_{\varepsilon_i}(x_i)$

then $A$ is open.

"$\Rightarrow$" Let $A$ open set. Let $\{V_{\alpha}\}$ such that $\alpha \in A$, a collection of sets such that $A\subset \bigcup_{\alpha\in A}V_{\alpha}$.

How I can get countable union?

Thanks for your help.

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    $\begingroup$ Use balls of radius $\frac 1k$ and center $\left(k_1+ j_12^{-l_1},\ldots,k_n+ j_n2^{-l_n}\right)$ where $k_i, j_i$ and $l_i$ are integers. $\endgroup$
    – Davide Giraudo
    Commented Dec 7, 2011 at 15:35
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    $\begingroup$ So, "collection of open balls accounting" means "countable collection of open balls"? $\endgroup$
    – David Mitra
    Commented Dec 7, 2011 at 15:58
  • $\begingroup$ How do you define open? I ask because if the definition is the usual one, then you do not give a proof for the right-to-left direction. The usual definition is that a set is open if and only if it is a union of open balls. It is true that if $A$ is open, then any point of $A$ is the center of an open ball contained in $A$ (which is what you are showing), but this is usually a theorem, not the definition of open. $\endgroup$
    – Andrés E. Caicedo
    Commented Dec 7, 2011 at 16:59
  • $\begingroup$ @David: yes, I'm sorry my english not good. "Countable collection of open balls" is correct. $\endgroup$
    – Hiperion
    Commented Dec 7, 2011 at 17:10
  • $\begingroup$ @Andres: $A\subset \mathbb{R}^n$ is open if for all $x\in A$ exists $\varepsilon>0$ such that $\mathbb{B}_{\varepsilon}(x)\subset A$ $\endgroup$
    – Hiperion
    Commented Dec 7, 2011 at 17:13

1 Answer 1

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Let $A\subset \mathbb R^n$ be open, then for each $x\in A$, $\mathbb B_\epsilon(x) \subset A$ some $\epsilon>0$,we can choose a rational number $q<\epsilon/2$ and a point $y$ with rational coordinates s.t. $||x-y||<q$. Then $x\in\mathbb B_q(y)\subset A$. The union of such $\mathbb B_q(y)$'s is $A$ and the collection is obviously countable.

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  • $\begingroup$ Thank you very much Andrew! $\endgroup$
    – Hiperion
    Commented Dec 7, 2011 at 17:14

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