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I know that every open set in $\mathbb{R}$ is a countable disjoint union of open intervals (possibly infinite). It is not difficult to prove it. For example, there is a proof in Real Analysis (Folland). I am sure if we do not require intervals to be disjoint, there should be a much simpler proof. This is what I am looking for.

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  • $\begingroup$ There is a much simpler proof if we do not require disjoint, but I don't know one which easily proves that the collection is countable. $\endgroup$
    – Arthur
    Mar 13, 2018 at 17:43
  • $\begingroup$ Yes, if we do not require countability, then the result follows directly from the definition of open set. $\endgroup$
    – Yerbolat
    Mar 13, 2018 at 17:49

1 Answer 1

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Let $G$ be an open set in ${\bf{R}}$. Let $x\in G$ and $r>0$ be such that $B_{2r}(x)\subseteq G$. Choose a rational number $p_{x}$ such that $|p_{x}-x|<r$. Choose a rational number $r_{x}>0$ such that $|p_{x}-x|<r_{x}<r$, then $x\in B_{r_{x}}(p_{x})$. Now we claim that $B_{r_{x}}(p_{x})\subseteq G$. For $y\in B_{r_{x}}(p_{x})$, then $|y-p_{x}|<r_{x}$, then $|y-x|\leq|y-p_{x}|+|p_{x}-x|<r_{x}+r<2r$, so $y\in B_{2r}(x)\subseteq G$, as expected.

So $G=\displaystyle\bigcup_{x\in G}B_{r_{x}}(p_{x})$ and all such $B_{r_{x}}(p_{x})$ are countably many.

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  • $\begingroup$ Thank you! It is very simple. $\endgroup$
    – Yerbolat
    Mar 13, 2018 at 18:07

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