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Can any subset of $\mathbb{R}$ be generated by taking countable unions, countable intersections and complements of open intervals?

Clearly, singletons can be generated from the complement of the union of half-rays, e.g.: $$a=((-\infty, a) \cup (a, \infty))^C .$$

Closed intervals can also be generated by a countable intersection of open sets of the form $(a- \frac1n, b+\frac1n)$ and similarly for half-open intervals.

From this, it seems obvious that any countable union/intersection of intervals can be generated.

How about for example uncountable unions/intersections of intervals? It is unclear to me whether this is enough to generate all subsets of $\mathbb{R}$. Can this be done?

Stephen
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    $\mathbb{R} \setminus \mathbb{Q}$ is the complement of $\Bbb Q$... – TonyK Dec 25 '19 at 13:48
  • @TonyK Of course it is :) I was too focused on trying to deal with the uncountability of the irrationals, and missed the obvious... – Stephen Dec 25 '19 at 13:58

2 Answers2

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Not every subset of $\mathbb{R}$ can be "created" by this process. If it were possible, we would get that the Borel $\sigma$-algebra would contain all subsets of $\mathbb{R}$, which it does not.

C_M
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    Thanks. Is there any simple example of a subset of $R$ that isn't a Borel set? – Stephen Dec 25 '19 at 14:20
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    Not really, all examples of non-Borel measurable sets depend on the axiom of choice (https://mathoverflow.net/questions/32720/non-borel-sets-without-axiom-of-choice). However, a short example is this one: https://math.stackexchange.com/questions/141017/lebesgue-measurable-set-that-is-not-a-borel-measurable-set – C_M Dec 25 '19 at 14:28
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______Editing my comment to extend the answer_______________

Hint: There is an example for Lebesgue but not Borel measurable sets.

Additionally:-

If you're familiar with measure theory, then there's a couple results that are somewhat related to this matter.

  1. Every non-empty open subset $U \subseteq \mathbb{R}$ is a disjoint union of open intervals.

  2. Littlewood's first principle of Analysis

Every measurable set $E\subset \mathbb{R}$ with $m(E)<$ $\infty$ is almost a finite union of intervals.

SL_MathGuy
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    I know some measure theory, but perhaps not enough to understand all of what you're saying. I'm not sure I understand what you mean by the hint. – Stephen Dec 25 '19 at 14:17