Q:

how many borel subsets of reals are there?

Can someone provide me the proof, or atleast a link to where I can find the proof of the fact: there are as many borel subsets of the reals, as there are real numbers? Of course, I am assuming AC.

A:

There are $2^{\aleph_0}$ many open sets. The Borel sets are defined as the union of a hierarchy of length $\aleph_1$. Thus we show that each stage of the hierarchy has size 
$2^{\aleph_0}$ so in total there will be $\aleph_1 2^{\aleph_0}= 2^{\aleph_0}$ many Borel
sets. Each successor stage of the hierarchy is formed from complements or from unions of countable subsets of the previous stage. But since the number of countable subsets of a set of size $2^{\aleph_0}$ has again size $2^{\aleph_0}$. We see that all levels have size $2^{\aleph_0}$.

