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Consider the smallest sigma algebra $\mathscr{B}$ generated by all open subsets of $\mathbb{R}$. One would expect that $\mathscr{B}$ contains all subsets of $\mathbb{R}$, but as it turns out, if we assume the axiom of choice to be true, there are some subsets of $\mathbb{R}$ which don't belong to $\mathscr{B}$. I was hoping if someone could help me out with a proof of the above statement, i.e, $\mathscr{B}$ does not contain all subsets of $\mathbb{R}$.

Asaf Karagila
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sayantankhan
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  • This is answered elsewhere on this site, but a short sketch is this: $\aleph_1=\omega_1$, the first uncountable cardinal, is regular (assuming choice), that is, countable unions of countable ordinals result in countable ordinals. This shows that $\mathcal B=\bigcup_{\alpha<\omega_1}\mathcal B_\alpha$, where each $\mathcal B_\alpha$ is the result of considering complements of the sets in the previous stages, and countable unions of such sets, with $\mathcal B_0$ being the collection of open sets. (Regularity is used to argue that this union $\bigcup_{\alpha<\omega_1}\mathcal B_\alpha$ (Cont.) – Andrés E. Caicedo Mar 27 '15 at 16:42
  • (cont) is indeed a $\sigma$-algebra). Now one argues by transfinite induction that each $\mathcal B_\alpha$ has the same size as $\mathbb R$, and Cantor's theorem gives us the result you are after. This argument assumes some familiarity with ordinals. One can avoid this by arguing that to each Borel set one can associate a certain "tagged tree", essentially a code that describes how from basic open sets we can by countable unions and complements reach the set in question. So one then only needs to count tagged trees, and there are only as many as there are reals. – Andrés E. Caicedo Mar 27 '15 at 16:46
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    "One would expect that $\mathscr{B}$ contains all subsets of $\mathbb{R}$": why on Earth would one expect that?! (Sorry for cutting the previous comment in half) – Najib Idrissi Mar 27 '15 at 16:46
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    @NajibIdrissi Probably because it is impossible to exhibit a set that is not Borel via an argument that does not invoke the axiom of choice. So, as long as everything one does is "explicit", only Borel sets are reached. – Andrés E. Caicedo Mar 27 '15 at 16:48
  • @AndresCaicedo: I don't think that is true. If you replace "Borel" by "Lebesgue", then you are right, but I the existence of non-Borel sets can be shown without choice (IIRC). EDIT: See also here: http://math.stackexchange.com/questions/286656/lebesgue-measure-borel-sets-and-axiom-of-choice. Under DC, one can show that there are non Borel sets. – PhoemueX Mar 27 '15 at 19:11
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    @PhoemueX It is true, I have addressed this topic a few times here and on Mathoverflow, so references should not be hard to find. Anyway, it is consistent with set theory without choice that $\mathbb R $ is a countable union of countable sets, in which case all sets of reals are Borel. – Andrés E. Caicedo Mar 27 '15 at 19:17
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    @AndresCaicedo: Yes, sorry, I just read that in the thread I linked. But at least there are (much) weaker forms of choice than AC (like DC) which imply that not all sets are Borel measurable. – PhoemueX Mar 27 '15 at 19:19

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