# Does Borel $\sigma$-algebra cardinality equal cardinality of the continuum?

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 $$\mathbb{R}$$?

• Yes..if i remember correctly it can be proved by transfinite induction – Marios Gretsas Dec 17 '19 at 18:24
• @ Marios Gretsas, may you give the link to the proof of this statement? – Даниил Владимирович Воробьев Dec 17 '19 at 18:27
• You can google it... i beleive that exists a proof in this site also – Marios Gretsas Dec 17 '19 at 18:28

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:
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.