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I know that cardinality of the $2^{\mathbb{R}}$ is greater than cardinality of $\mathbb{R}$. Does the cardinality of the Borel $\sigma$-algebra equal the cardinality of the $\mathbb{R}$?

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  • $\begingroup$ Yes..if i remember correctly it can be proved by transfinite induction $\endgroup$ – Marios Gretsas Dec 17 '19 at 18:24
  • $\begingroup$ @ Marios Gretsas, may you give the link to the proof of this statement? $\endgroup$ – Даниил Владимирович Воробьев Dec 17 '19 at 18:27
  • $\begingroup$ You can google it... i beleive that exists a proof in this site also $\endgroup$ – Marios Gretsas Dec 17 '19 at 18:28
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The standard proof of this argument uses that the Borel sets can be produced through a transfinite process of length $\omega_1$: start with the open sets, iterate the process of taking complements and countable unions. One argues inductively that there are $\mathfrak c$ sets at each stage, and since $\aleph_1$ is regular and less than or equal to $\mathfrak c$, the equality follows.

One may want a proof that avoids any mention of $\omega_1$. This is possible working with codes (so you count codes for Borel sets rather than the Borel sets themselves; the "codes" are a way of keeping track of how the Borel set came to be starting from basic open sets). I sketched an argument in another answer on this site, or you can see a recent paper with details along the same lines:

MR4003698. Kánnai, Z. An elementary proof that the Borel class of the reals has cardinality continuum. Acta Math. Hungar. 159 (2019), no. 1, 124–130.

Note that all these arguments by necessity use some form of the axiom of choice (eiher by arguing that $\aleph_1\le\mathfrak c$, or that a quotient of $\mathbb R$ has at most the same size as $\mathbb R$), since it is consistent with the failure of choice that the reals are a countable union of countable sets and therefore all sets of reals are Borel.

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