# Open set as a countable union of spheres

Could you tell me how to prove that each open set can is a countable union of open balls: $K\left( x,r \right) = \left\{ y \in \mathbb{R} ^{n} | \ d(x,y) <r \right\}$ where d is the Euclid metric?

I know that $U$ is an open set if $\ int \ U=U$

and for $E \subset \mathbb{R}, \ \$ $intE=\bigcup \left\{K \ | \ K\subset E \ \right\}$ (K as defined above).

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I think your terminology is rather unusual. What you call a sphere is usually know as an open ball and its boundary is what is conventionally known as spheres (in $\mathbb{R}^n$). Also, nowaday people call what you call a sum a union. Under the usual interpretation, no nonempty open set is a union of spheres. – Michael Greinecker Jan 11 '13 at 11:46

$U$ is open, so for each $x \in U$, there exists $r > 0$ such that $K(x,r) \subset U$. Now you can find rational elements $p,q$ ($p \in \mathbb Q^n, q \in \mathbb Q$) such that $K(p,q) \subset K(x,r)$ and $x \in K(p,q)$.

This means that you can express $U$ as the union of such $K(p,q)$. And there are only countably many rational pairs $(p,q)$.

(This is basically the same as Pambos's answer, it just uses more words.)

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Yes is basically the same but I like yours better (too) :) – P.. Jan 11 '13 at 11:58

Prove that $U\cap \mathbb Q^n$ is dense in $U$.
Then consider the family $\mathcal A =\left\{K(z,q):z\in U\cap \mathbb Q^n, q \in \mathbb Q,K(z,q)\subseteq U\right\}$.
Prove that $\mathcal A$ is countable and $\cup \mathcal A=U.$

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