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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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    @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$2011-12-07

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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||. 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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    Thank you very much $A$ndrew!2011-12-07