# Borel sigma algebra not containing all subsets of $\mathbb{R}$?

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 $\mathbb{R}$.

• 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.) Mar 27, 2015 at 16:42
• "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) Mar 27, 2015 at 16:46
• @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. Mar 27, 2015 at 16:48
• @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. Mar 27, 2015 at 19:17
• @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. Mar 27, 2015 at 19:19