# set theory – How many Non-Borel Sets are there?

I’m currently doing a course on measure theoretic probability. In the course I learned that the cardinality of Borel $$sigma$$-algebra $$mathscr{B}$$ is the same as that of continuum $$mathbb{R}$$. But what is the cardinality of the set of Non-Borel Sets, let’s call it $$mathcal{S}$$.

My inital guess is along these lines. That since $$mathcal{S} = mathscr{B^{c}}$$. We can write that,
$$mathbb{2^R} = mathcal{S} ~ cup mathscr{B}$$ Assume that $$mathcal{S}$$ has the same cardinality as $$mathbb{R}$$. Now we know that $$mathscr{B}$$ also has the same cardinality as that of $$mathbb{R}$$. But their union $$mathbb{2^R}$$ has a stricly greater cardinality than $$mathbb{R}$$. Again assuming that union of uncountable sets of same cardinality has the same cardinality (I’m aware that this is true for countable sets, but not sure about uncountables), this results in a contradiction.

Hence the set $$mathcal{S}$$ must have a strictly bigger cardinality than $$mathbb{R}$$. Is this correct?