Let $O\subset \mathbb R^2$ an open and let denote $B(x,r)$ the ball centred in $x$ with radius $r$. I want to show that $O$ is a countable union of $B(x,r)$. Let $x\in O$. Then, there is $r$ s.t. $B(x,r)\subset O$. I'm sure that by density of $\mathbb Q^2$ in $\mathbb R^2$, I can show that there is $r'\in\mathbb Q$ and $y\in O\cap \mathbb Q^2$ s.t. $$B(x,r)\subset B(y,r')\subset O,$$ but I really have difficulties to construct $B(y,r')$. Any help would be welcome.
Show that if $O\subset \mathbb R^2$ open then $O=\bigcup_{countable}B(x,r)$
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general-topology
1 Answers
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Let $B(x,r/4)\subset B(x,r)$. Then, take $y\in \mathbb Q^2\cap B(x,r/4)$. In particular, $$B(x,r/4)\subset B(y,r/2)\subset B(x,r)\subset O.$$