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We only consider Borel sets on $\mathbb R$. As we know, the Borel $\sigma$-algebra is constructed transfinitely as follows:

  1. Let $B_0=\mathcal T$ be the set of open sets on $\mathbb R$;
  2. If $\alpha$ is a limit ordinal, $B_\alpha=\bigcup_{\beta<\alpha}B_\beta$;
  3. Otherwise, let $\beta=\alpha-1$, and $B_\alpha=(B_\beta)_{\delta\sigma}$, where $A_\delta$ denotes the collection of countable intersections of sets in $A$, and $A_{\delta\sigma}$ denotes the collection of countable unions of sets in $A_\delta$;
  4. Let $\mathcal B=B_{\omega_1}$ where $\omega_1$ is the smallest uncountable ordinal, then $\mathcal B$ is the Borel $\sigma$-algebra.

However, I wonder whether $\omega_1$ is the smallest ordinal $\alpha$ such that $B_\alpha=\mathcal B$, i.e. the minimal step to finish the process. As we know, $((0,1)\cap\mathbb Q)\cup([2,3]\setminus\mathbb Q)$ is a Borel set which is neither $F_\sigma$ nor $G_\delta$.

Any idea? Thanks!

Yai0Phah
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    Yes, $\omega_1$ steps are needed. One way of seeing this is to note that the Baire hierarchy of functions requires $\omega_1$ steps (see here), and each level can be characterized in terms of the Borel complexity of the preimages of open sets. One can also argue directly, in terms of universal sets. See Kechris's Classical descriptive set theory. On the other hand, without the axiom of choice, it is consistent both that the hierarchy is longer, and that it is shorter (see here). – Andrés E. Caicedo Apr 08 '14 at 00:16

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