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Let $E$ be a nonempty open subset of $\Bbb R$. There is a collection $A$ of subsets of $\Bbb R$ satisfying the following:

  • $(i)$ Each element of $A$ is an open interval (possibly infinite) and any two distinct members of $A$ are disjoint;
  • $(ii)$ $A$ is at most countable;
  • $(iii)$ $E =\bigcup\limits_{G\in A}G$.

Prove that ”Any open subset of $\Bbb R$ is an at most countable union of pairwise disjoint open intervals”.

I don't know any structure theorems. I only have done a course on real analysis. Which dealt with compact sets and all.

bof
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ramAN
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1 Answers1

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Let $\mathcal{I}$ be the set of the intervals in $\mathbb{R}$ with rational endpoints and $E \subset \mathbb{R}$ an open set.

  1. For all $x \in E$, prove there exist a rational $r_x > 0$ and a rational $q_x$ such that $x \in\, ]r_x-q_x, q_x + r_x[$ and $]r_x-q_x, q_x + r_x[ \subset E$. Hint: use the density of $\mathbb{Q}$ in $\mathbb{R}$.
  2. Deduce that $E$ can be written as a union of elements of $\mathcal{I}$.
  3. Prove that $\mathcal{I}$ is countable. Hint: use the fact that $\mathbb{Q}\times \mathbb{Q}$ is countable.
  4. Deduce you can write $E$ as a countable union of elements of $\mathcal{I}$.
  5. Let $E = \cup_{\alpha \in A} I_{\alpha}$ be such a union. For each $\alpha \in A$, prove there exists a largest open interval $J_{\alpha}$ such that $I_{\alpha} \subset J_{\alpha} \subset E$. This $J_{\alpha}$ is not necessary in $\mathcal{I}$.
  6. Conclude you can write $E$ as a countable union of pairwise disjoint intervals.
halbaroth
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